/*

  * Copyright 2009 Google Inc.  All Rights Reserved.

  * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.

  *

  * This code is free software; you can redistribute it and/or modify it

  * under the terms of the GNU General Public License version 2 only, as

  * published by the Free Software Foundation.  Oracle designates this

  * particular file as subject to the "Classpath" exception as provided

  * by Oracle in the LICENSE file that accompanied this code.

  *

  * This code is distributed in the hope that it will be useful, but WITHOUT

  * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or

  * FITNESS FOR A PARTICULAR PURPOSE.  See the GNU General Public License

  * version 2 for more details (a copy is included in the LICENSE file that

  * accompanied this code).

  *

  * You should have received a copy of the GNU General Public License version

  * 2 along with this work; if not, write to the Free Software Foundation,

  * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.

  *

  * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA

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  */

 package java.util;

 /**

  * A stable, adaptive, iterative mergesort that requires far fewer than

  * n lg(n) comparisons when running on partially sorted arrays, while

  * offering performance comparable to a traditional mergesort when run

  * on random arrays.  Like all proper mergesorts, this sort is stable and

  * runs O(n log n) time (worst case).  In the worst case, this sort requires

  * temporary storage space for n/2 object references; in the best case,

  * it requires only a small constant amount of space.

  *

  * This implementation was adapted from Tim Peters's list sort for

  * Python, which is described in detail here:

  *

  *   http://svn.python.org/projects/python/trunk/Objects/listsort.txt

  *

  * Tim's C code may be found here:

  *

  *   http://svn.python.org/projects/python/trunk/Objects/listobject.c

  *

  * The underlying techniques are described in this paper (and may have

  * even earlier origins):

  *

  *  "Optimistic Sorting and Information Theoretic Complexity"

  *  Peter McIlroy

  *  SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms),

  *  pp 467-474, Austin, Texas, 25-27 January 1993.

  *

  * While the API to this class consists solely of static methods, it is

  * (privately) instantiable; a TimSort instance holds the state of an ongoing

  * sort, assuming the input array is large enough to warrant the full-blown

  * TimSort. Small arrays are sorted in place, using a binary insertion sort.

  *

  * @author Josh Bloch

  */

 class TimSort<T> {

     /**

      * This is the minimum sized sequence that will be merged.  Shorter

      * sequences will be lengthened by calling binarySort.  If the entire

      * array is less than this length, no merges will be performed.

      *

      * This constant should be a power of two.  It was 64 in Tim Peter's C

      * implementation, but 32 was empirically determined to work better in

      * this implementation.  In the unlikely event that you set this constant

      * to be a number that's not a power of two, you'll need to change the

      * {@link #minRunLength} computation.

      *

      * If you decrease this constant, you must change the stackLen

      * computation in the TimSort constructor, or you risk an

      * ArrayOutOfBounds exception.  See listsort.txt for a discussion

      * of the minimum stack length required as a function of the length

      * of the array being sorted and the minimum merge sequence length.

      */

     private static final int MIN_MERGE = 32;

     /**

      * The array being sorted.

      */

     private final T[] a;

     /**

      * The comparator for this sort.

      */

     private final Comparator<? super T> c;

     /**

      * When we get into galloping mode, we stay there until both runs win less

      * often than MIN_GALLOP consecutive times.

      */

     private static final int  MIN_GALLOP = 7;

     /**

      * This controls when we get *into* galloping mode.  It is initialized

      * to MIN_GALLOP.  The mergeLo and mergeHi methods nudge it higher for

      * random data, and lower for highly structured data.

      */

     private int minGallop = MIN_GALLOP;

     /**

      * Maximum initial size of tmp array, which is used for merging.  The array

      * can grow to accommodate demand.

      *

      * Unlike Tim's original C version, we do not allocate this much storage

      * when sorting smaller arrays.  This change was required for performance.

      */

     private static final int INITIAL_TMP_STORAGE_LENGTH = 256;

     /**

      * Temp storage for merges.

      */

     private T[] tmp; // Actual runtime type will be Object[], regardless of T

     /**

      * A stack of pending runs yet to be merged.  Run i starts at

      * address base[i] and extends for len[i] elements.  It's always

      * true (so long as the indices are in bounds) that:

      *

      *     runBase[i] + runLen[i] == runBase[i + 1]

      *

      * so we could cut the storage for this, but it's a minor amount,

      * and keeping all the info explicit simplifies the code.

      */

     private int stackSize = 0;  // Number of pending runs on stack

     private final int[] runBase;

     private final int[] runLen;

     /**

      * Creates a TimSort instance to maintain the state of an ongoing sort.

      *

      * @param a the array to be sorted

      * @param c the comparator to determine the order of the sort

      */

     private TimSort(T[] a, Comparator<? super T> c) {

         this.a = a;

         this.c = c;

         // Allocate temp storage (which may be increased later if necessary)

         int len = a.length;

         @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})

         T[] newArray = (T[]) new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ?

                                         len >>> 1 : INITIAL_TMP_STORAGE_LENGTH];

         tmp = newArray;

         /*

          * Allocate runs-to-be-merged stack (which cannot be expanded).  The

          * stack length requirements are described in listsort.txt.  The C

          * version always uses the same stack length (85), but this was

          * measured to be too expensive when sorting "mid-sized" arrays (e.g.,

          * 100 elements) in Java.  Therefore, we use smaller (but sufficiently

          * large) stack lengths for smaller arrays.  The "magic numbers" in the

          * computation below must be changed if MIN_MERGE is decreased.  See

          * the MIN_MERGE declaration above for more information.

          */

         int stackLen = (len <    120  ?  5 :

                         len <   1542  ? 10 :

                         len < 119151  ? 19 : 40);

         runBase = new int[stackLen];

         runLen = new int[stackLen];

     }

     /*

      * The next two methods (which are package private and static) constitute

      * the entire API of this class.  Each of these methods obeys the contract

      * of the public method with the same signature in java.util.Arrays.

      */

     static <T> void sort(T[] a, Comparator<? super T> c) {

         sort(a, 0, a.length, c);

     }

     static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c) {

         if (c == null) {

             Arrays.sort(a, lo, hi);

             return;

         }

         rangeCheck(a.length, lo, hi);

         int nRemaining  = hi - lo;

         if (nRemaining < 2)

             return;  // Arrays of size 0 and 1 are always sorted

         // If array is small, do a "mini-TimSort" with no merges

         if (nRemaining < MIN_MERGE) {

             int initRunLen = countRunAndMakeAscending(a, lo, hi, c);

             binarySort(a, lo, hi, lo + initRunLen, c);

             return;

         }

         /**

          * March over the array once, left to right, finding natural runs,

          * extending short natural runs to minRun elements, and merging runs

          * to maintain stack invariant.

          */

         TimSort<T> ts = new TimSort<>(a, c);

         int minRun = minRunLength(nRemaining);

         do {

             // Identify next run

             int runLen = countRunAndMakeAscending(a, lo, hi, c);

             // If run is short, extend to min(minRun, nRemaining)

             if (runLen < minRun) {

                 int force = nRemaining <= minRun ? nRemaining : minRun;

                 binarySort(a, lo, lo + force, lo + runLen, c);

                 runLen = force;

             }

             // Push run onto pending-run stack, and maybe merge

             ts.pushRun(lo, runLen);

             ts.mergeCollapse();

             // Advance to find next run

             lo += runLen;

             nRemaining -= runLen;

         } while (nRemaining != 0);

         // Merge all remaining runs to complete sort

         assert lo == hi;

         ts.mergeForceCollapse();

         assert ts.stackSize == 1;

     }

     /**

      * Sorts the specified portion of the specified array using a binary

      * insertion sort.  This is the best method for sorting small numbers

      * of elements.  It requires O(n log n) compares, but O(n^2) data

      * movement (worst case).

      *

      * If the initial part of the specified range is already sorted,

      * this method can take advantage of it: the method assumes that the

      * elements from index {@code lo}, inclusive, to {@code start},

      * exclusive are already sorted.

      *

      * @param a the array in which a range is to be sorted

      * @param lo the index of the first element in the range to be sorted

      * @param hi the index after the last element in the range to be sorted

      * @param start the index of the first element in the range that is

      *        not already known to be sorted ({@code lo <= start <= hi})

      * @param c comparator to used for the sort

      */

     @SuppressWarnings("fallthrough")

     private static <T> void binarySort(T[] a, int lo, int hi, int start,

                                        Comparator<? super T> c) {

         assert lo <= start && start <= hi;

         if (start == lo)

             start++;

         for ( ; start < hi; start++) {

             T pivot = a[start];

             // Set left (and right) to the index where a[start] (pivot) belongs

             int left = lo;

             int right = start;

             assert left <= right;

             /*

              * Invariants:

              *   pivot >= all in [lo, left).

              *   pivot <  all in [right, start).

              */

             while (left < right) {

                 int mid = (left + right) >>> 1;

                 if (c.compare(pivot, a[mid]) < 0)

                     right = mid;

                 else

                     left = mid + 1;

             }

             assert left == right;

             /*

              * The invariants still hold: pivot >= all in [lo, left) and

              * pivot < all in [left, start), so pivot belongs at left.  Note

              * that if there are elements equal to pivot, left points to the

              * first slot after them -- that's why this sort is stable.

              * Slide elements over to make room for pivot.

              */

             int n = start - left;  // The number of elements to move

             // Switch is just an optimization for arraycopy in default case

             switch (n) {

                 case 2:  a[left + 2] = a[left + 1];

                 case 1:  a[left + 1] = a[left];

                          break;

                 default: System.arraycopy(a, left, a, left + 1, n);

             }

             a[left] = pivot;

         }

     }

     /**

      * Returns the length of the run beginning at the specified position in

      * the specified array and reverses the run if it is descending (ensuring

      * that the run will always be ascending when the method returns).

      *

      * A run is the longest ascending sequence with:

      *

      *    a[lo] <= a[lo + 1] <= a[lo + 2] <= ...

      *

      * or the longest descending sequence with:

      *

      *    a[lo] >  a[lo + 1] >  a[lo + 2] >  ...

      *

      * For its intended use in a stable mergesort, the strictness of the

      * definition of "descending" is needed so that the call can safely

      * reverse a descending sequence without violating stability.

      *

      * @param a the array in which a run is to be counted and possibly reversed

      * @param lo index of the first element in the run

      * @param hi index after the last element that may be contained in the run.

               It is required that {@code lo < hi}.

      * @param c the comparator to used for the sort

      * @return  the length of the run beginning at the specified position in

      *          the specified array

      */

     private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi,

                                                     Comparator<? super T> c) {

         assert lo < hi;

         int runHi = lo + 1;

         if (runHi == hi)

             return 1;

         // Find end of run, and reverse range if descending

         if (c.compare(a[runHi++], a[lo]) < 0) { // Descending

             while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)

                 runHi++;

             reverseRange(a, lo, runHi);

         } else {                              // Ascending

             while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)

                 runHi++;

         }

         return runHi - lo;

     }

     /**

      * Reverse the specified range of the specified array.

      *

      * @param a the array in which a range is to be reversed

      * @param lo the index of the first element in the range to be reversed

      * @param hi the index after the last element in the range to be reversed

      */

     private static void reverseRange(Object[] a, int lo, int hi) {

         hi--;

         while (lo < hi) {

             Object t = a[lo];

             a[lo++] = a[hi];

             a[hi--] = t;

         }

     }

     /**

      * Returns the minimum acceptable run length for an array of the specified

      * length. Natural runs shorter than this will be extended with

      * {@link #binarySort}.

      *

      * Roughly speaking, the computation is:

      *

      *  If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).

      *  Else if n is an exact power of 2, return MIN_MERGE/2.

      *  Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k

      *   is close to, but strictly less than, an exact power of 2.

      *

      * For the rationale, see listsort.txt.

      *

      * @param n the length of the array to be sorted

      * @return the length of the minimum run to be merged

      */

     private static int minRunLength(int n) {

         assert n >= 0;

         int r = 0;      // Becomes 1 if any 1 bits are shifted off

         while (n >= MIN_MERGE) {

             r |= (n & 1);

             n >>= 1;

         }

         return n + r;

     }

     /**

      * Pushes the specified run onto the pending-run stack.

      *

      * @param runBase index of the first element in the run

      * @param runLen  the number of elements in the run

      */

     private void pushRun(int runBase, int runLen) {

         this.runBase[stackSize] = runBase;

         this.runLen[stackSize] = runLen;

         stackSize++;

     }

     /**

      * Examines the stack of runs waiting to be merged and merges adjacent runs

      * until the stack invariants are reestablished:

      *

      *     1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]

      *     2. runLen[i - 2] > runLen[i - 1]

      *

      * This method is called each time a new run is pushed onto the stack,

      * so the invariants are guaranteed to hold for i < stackSize upon

      * entry to the method.

      */

     private void mergeCollapse() {

         while (stackSize > 1) {

             int n = stackSize - 2;

             if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {

                 if (runLen[n - 1] < runLen[n + 1])

                     n--;

                 mergeAt(n);

             } else if (runLen[n] <= runLen[n + 1]) {

                 mergeAt(n);

             } else {

                 break; // Invariant is established

             }

         }

     }

     /**

      * Merges all runs on the stack until only one remains.  This method is

      * called once, to complete the sort.

      */

     private void mergeForceCollapse() {

         while (stackSize > 1) {

             int n = stackSize - 2;

             if (n > 0 && runLen[n - 1] < runLen[n + 1])

                 n--;

             mergeAt(n);

         }

     }

     /**

      * Merges the two runs at stack indices i and i+1.  Run i must be

      * the penultimate or antepenultimate run on the stack.  In other words,

      * i must be equal to stackSize-2 or stackSize-3.

      *

      * @param i stack index of the first of the two runs to merge

      */

     private void mergeAt(int i) {

         assert stackSize >= 2;

         assert i >= 0;

         assert i == stackSize - 2 || i == stackSize - 3;

         int base1 = runBase[i];

         int len1 = runLen[i];

         int base2 = runBase[i + 1];

         int len2 = runLen[i + 1];

         assert len1 > 0 && len2 > 0;

         assert base1 + len1 == base2;

         /*

          * Record the length of the combined runs; if i is the 3rd-last

          * run now, also slide over the last run (which isn't involved

          * in this merge).  The current run (i+1) goes away in any case.

          */

         runLen[i] = len1 + len2;

         if (i == stackSize - 3) {

             runBase[i + 1] = runBase[i + 2];

             runLen[i + 1] = runLen[i + 2];

         }

         stackSize--;

         /*

          * Find where the first element of run2 goes in run1. Prior elements

          * in run1 can be ignored (because they're already in place).

          */

         int k = gallopRight(a[base2], a, base1, len1, 0, c);

         assert k >= 0;

         base1 += k;

         len1 -= k;

         if (len1 == 0)

             return;

         /*

          * Find where the last element of run1 goes in run2. Subsequent elements

          * in run2 can be ignored (because they're already in place).

          */

         len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);

         assert len2 >= 0;

         if (len2 == 0)

             return;

         // Merge remaining runs, using tmp array with min(len1, len2) elements

         if (len1 <= len2)

             mergeLo(base1, len1, base2, len2);

         else

             mergeHi(base1, len1, base2, len2);

     }

     /**

      * Locates the position at which to insert the specified key into the

      * specified sorted range; if the range contains an element equal to key,

      * returns the index of the leftmost equal element.

      *

      * @param key the key whose insertion point to search for

      * @param a the array in which to search

      * @param base the index of the first element in the range

      * @param len the length of the range; must be > 0

      * @param hint the index at which to begin the search, 0 <= hint < n.

      *     The closer hint is to the result, the faster this method will run.

      * @param c the comparator used to order the range, and to search

      * @return the int k,  0 <= k <= n such that a[b + k - 1] < key <= a[b + k],

      *    pretending that a[b - 1] is minus infinity and a[b + n] is infinity.

      *    In other words, key belongs at index b + k; or in other words,

      *    the first k elements of a should precede key, and the last n - k

      *    should follow it.

      */

     private static <T> int gallopLeft(T key, T[] a, int base, int len, int hint,

                                       Comparator<? super T> c) {

         assert len > 0 && hint >= 0 && hint < len;

         int lastOfs = 0;

         int ofs = 1;

         if (c.compare(key, a[base + hint]) > 0) {

             // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]

             int maxOfs = len - hint;

             while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) {

                 lastOfs = ofs;

                 ofs = (ofs << 1) + 1;

                 if (ofs <= 0)   // int overflow

                     ofs = maxOfs;

             }

             if (ofs > maxOfs)

                 ofs = maxOfs;

             // Make offsets relative to base

             lastOfs += hint;

             ofs += hint;

         } else { // key <= a[base + hint]

             // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]

             final int maxOfs = hint + 1;

             while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) {

                 lastOfs = ofs;

                 ofs = (ofs << 1) + 1;

                 if (ofs <= 0)   // int overflow

                     ofs = maxOfs;

             }

             if (ofs > maxOfs)

                 ofs = maxOfs;

             // Make offsets relative to base

             int tmp = lastOfs;

             lastOfs = hint - ofs;

             ofs = hint - tmp;

         }

         assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

         /*

          * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere

          * to the right of lastOfs but no farther right than ofs.  Do a binary

          * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].

          */

         lastOfs++;

         while (lastOfs < ofs) {

             int m = lastOfs + ((ofs - lastOfs) >>> 1);

             if (c.compare(key, a[base + m]) > 0)

                 lastOfs = m + 1;  // a[base + m] < key

             else

                 ofs = m;          // key <= a[base + m]

         }

         assert lastOfs == ofs;    // so a[base + ofs - 1] < key <= a[base + ofs]

         return ofs;

     }

     /**

      * Like gallopLeft, except that if the range contains an element equal to

      * key, gallopRight returns the index after the rightmost equal element.

      *

      * @param key the key whose insertion point to search for

      * @param a the array in which to search

      * @param base the index of the first element in the range

      * @param len the length of the range; must be > 0

      * @param hint the index at which to begin the search, 0 <= hint < n.

      *     The closer hint is to the result, the faster this method will run.

      * @param c the comparator used to order the range, and to search

      * @return the int k,  0 <= k <= n such that a[b + k - 1] <= key < a[b + k]

      */

     private static <T> int gallopRight(T key, T[] a, int base, int len,

                                        int hint, Comparator<? super T> c) {

         assert len > 0 && hint >= 0 && hint < len;

         int ofs = 1;

         int lastOfs = 0;

         if (c.compare(key, a[base + hint]) < 0) {

             // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]

             int maxOfs = hint + 1;

             while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) {

                 lastOfs = ofs;

                 ofs = (ofs << 1) + 1;

                 if (ofs <= 0)   // int overflow

                     ofs = maxOfs;

             }

             if (ofs > maxOfs)

                 ofs = maxOfs;

             // Make offsets relative to b

             int tmp = lastOfs;

             lastOfs = hint - ofs;

             ofs = hint - tmp;

         } else { // a[b + hint] <= key

             // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]

             int maxOfs = len - hint;

             while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) {

                 lastOfs = ofs;

                 ofs = (ofs << 1) + 1;

                 if (ofs <= 0)   // int overflow

                     ofs = maxOfs;

             }

             if (ofs > maxOfs)

                 ofs = maxOfs;

             // Make offsets relative to b

             lastOfs += hint;

             ofs += hint;

         }

         assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

         /*

          * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to

          * the right of lastOfs but no farther right than ofs.  Do a binary

          * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].

          */

         lastOfs++;

         while (lastOfs < ofs) {

             int m = lastOfs + ((ofs - lastOfs) >>> 1);

             if (c.compare(key, a[base + m]) < 0)

                 ofs = m;          // key < a[b + m]

             else

                 lastOfs = m + 1;  // a[b + m] <= key

         }

         assert lastOfs == ofs;    // so a[b + ofs - 1] <= key < a[b + ofs]

         return ofs;

     }

     /**

      * Merges two adjacent runs in place, in a stable fashion.  The first

      * element of the first run must be greater than the first element of the

      * second run (a[base1] > a[base2]), and the last element of the first run

      * (a[base1 + len1-1]) must be greater than all elements of the second run.

      *

      * For performance, this method should be called only when len1 <= len2;

      * its twin, mergeHi should be called if len1 >= len2.  (Either method

      * may be called if len1 == len2.)

      *

      * @param base1 index of first element in first run to be merged

      * @param len1  length of first run to be merged (must be > 0)

      * @param base2 index of first element in second run to be merged

      *        (must be aBase + aLen)

      * @param len2  length of second run to be merged (must be > 0)

      */

     private void mergeLo(int base1, int len1, int base2, int len2) {

         assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

         // Copy first run into temp array

         T[] a = this.a; // For performance

         T[] tmp = ensureCapacity(len1);

         System.arraycopy(a, base1, tmp, 0, len1);

         int cursor1 = 0;       // Indexes into tmp array

         int cursor2 = base2;   // Indexes int a

         int dest = base1;      // Indexes int a

         // Move first element of second run and deal with degenerate cases

         a[dest++] = a[cursor2++];

         if (--len2 == 0) {

             System.arraycopy(tmp, cursor1, a, dest, len1);

             return;

         }

         if (len1 == 1) {

             System.arraycopy(a, cursor2, a, dest, len2);

             a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge

             return;

         }

         Comparator<? super T> c = this.c;  // Use local variable for performance

         int minGallop = this.minGallop;    //  "    "       "     "      "

     outer:

         while (true) {

             int count1 = 0; // Number of times in a row that first run won

             int count2 = 0; // Number of times in a row that second run won

             /*

              * Do the straightforward thing until (if ever) one run starts

              * winning consistently.

              */

             do {

                 assert len1 > 1 && len2 > 0;

                 if (c.compare(a[cursor2], tmp[cursor1]) < 0) {

                     a[dest++] = a[cursor2++];

                     count2++;

                     count1 = 0;

                     if (--len2 == 0)

                         break outer;

                 } else {

                     a[dest++] = tmp[cursor1++];

                     count1++;

                     count2 = 0;

                     if (--len1 == 1)

                         break outer;

                 }

             } while ((count1 | count2) < minGallop);

             /*

              * One run is winning so consistently that galloping may be a

              * huge win. So try that, and continue galloping until (if ever)

              * neither run appears to be winning consistently anymore.

              */

             do {

                 assert len1 > 1 && len2 > 0;

                 count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);

                 if (count1 != 0) {

                     System.arraycopy(tmp, cursor1, a, dest, count1);

                     dest += count1;

                     cursor1 += count1;

                     len1 -= count1;

                     if (len1 <= 1) // len1 == 1 || len1 == 0

                         break outer;

                 }

                 a[dest++] = a[cursor2++];

                 if (--len2 == 0)

                     break outer;

                 count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);

                 if (count2 != 0) {

                     System.arraycopy(a, cursor2, a, dest, count2);

                     dest += count2;

                     cursor2 += count2;

                     len2 -= count2;

                     if (len2 == 0)

                         break outer;

                 }

                 a[dest++] = tmp[cursor1++];

                 if (--len1 == 1)

                     break outer;

                 minGallop--;

             } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);

             if (minGallop < 0)

                 minGallop = 0;

             minGallop += 2;  // Penalize for leaving gallop mode

         }  // End of "outer" loop

         this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field

         if (len1 == 1) {

             assert len2 > 0;

             System.arraycopy(a, cursor2, a, dest, len2);

             a[dest + len2] = tmp[cursor1]; //  Last elt of run 1 to end of merge

         } else if (len1 == 0) {

             throw new IllegalArgumentException(

                 "Comparison method violates its general contract!");

         } else {

             assert len2 == 0;

             assert len1 > 1;

             System.arraycopy(tmp, cursor1, a, dest, len1);

         }

     }

     /**

      * Like mergeLo, except that this method should be called only if

      * len1 >= len2; mergeLo should be called if len1 <= len2.  (Either method

      * may be called if len1 == len2.)

      *

      * @param base1 index of first element in first run to be merged

      * @param len1  length of first run to be merged (must be > 0)

      * @param base2 index of first element in second run to be merged

      *        (must be aBase + aLen)

      * @param len2  length of second run to be merged (must be > 0)

      */

     private void mergeHi(int base1, int len1, int base2, int len2) {

         assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

         // Copy second run into temp array

         T[] a = this.a; // For performance

         T[] tmp = ensureCapacity(len2);

         System.arraycopy(a, base2, tmp, 0, len2);

         int cursor1 = base1 + len1 - 1;  // Indexes into a

         int cursor2 = len2 - 1;          // Indexes into tmp array

         int dest = base2 + len2 - 1;     // Indexes into a

         // Move last element of first run and deal with degenerate cases

         a[dest--] = a[cursor1--];

         if (--len1 == 0) {

             System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);

             return;

         }

         if (len2 == 1) {

             dest -= len1;

             cursor1 -= len1;

             System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);

             a[dest] = tmp[cursor2];

             return;

         }

         Comparator<? super T> c = this.c;  // Use local variable for performance

         int minGallop = this.minGallop;    //  "    "       "     "      "

     outer:

         while (true) {

             int count1 = 0; // Number of times in a row that first run won

             int count2 = 0; // Number of times in a row that second run won

             /*

              * Do the straightforward thing until (if ever) one run

              * appears to win consistently.

              */

             do {

                 assert len1 > 0 && len2 > 1;

                 if (c.compare(tmp[cursor2], a[cursor1]) < 0) {

                     a[dest--] = a[cursor1--];

                     count1++;

                     count2 = 0;

                     if (--len1 == 0)

                         break outer;

                 } else {

                     a[dest--] = tmp[cursor2--];

                     count2++;

                     count1 = 0;

                     if (--len2 == 1)

                         break outer;

                 }

             } while ((count1 | count2) < minGallop);

             /*

              * One run is winning so consistently that galloping may be a

              * huge win. So try that, and continue galloping until (if ever)

              * neither run appears to be winning consistently anymore.

              */

             do {

                 assert len1 > 0 && len2 > 1;

                 count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c);

                 if (count1 != 0) {

                     dest -= count1;

                     cursor1 -= count1;

                     len1 -= count1;

                     System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);

                     if (len1 == 0)

                         break outer;

                 }

                 a[dest--] = tmp[cursor2--];

                 if (--len2 == 1)

                     break outer;

                 count2 = len2 - gallopLeft(a[cursor1], tmp, 0, len2, len2 - 1, c);

                 if (count2 != 0) {

                     dest -= count2;

                     cursor2 -= count2;

                     len2 -= count2;

                     System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);

                     if (len2 <= 1)  // len2 == 1 || len2 == 0

                         break outer;

                 }

                 a[dest--] = a[cursor1--];

                 if (--len1 == 0)

                     break outer;

                 minGallop--;

             } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);

             if (minGallop < 0)

                 minGallop = 0;

             minGallop += 2;  // Penalize for leaving gallop mode

         }  // End of "outer" loop

         this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field

         if (len2 == 1) {

             assert len1 > 0;

             dest -= len1;

             cursor1 -= len1;

             System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);

             a[dest] = tmp[cursor2];  // Move first elt of run2 to front of merge

         } else if (len2 == 0) {

             throw new IllegalArgumentException(

                 "Comparison method violates its general contract!");

         } else {

             assert len1 == 0;

             assert len2 > 0;

             System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);

         }

     }

     /**

      * Ensures that the external array tmp has at least the specified

      * number of elements, increasing its size if necessary.  The size

      * increases exponentially to ensure amortized linear time complexity.

      *

      * @param minCapacity the minimum required capacity of the tmp array

      * @return tmp, whether or not it grew

      */

     private T[] ensureCapacity(int minCapacity) {

         if (tmp.length < minCapacity) {

             // Compute smallest power of 2 > minCapacity

             int newSize = minCapacity;

             newSize |= newSize >> 1;

             newSize |= newSize >> 2;

             newSize |= newSize >> 4;

             newSize |= newSize >> 8;

             newSize |= newSize >> 16;

             newSize++;

             if (newSize < 0) // Not bloody likely!

                 newSize = minCapacity;

             else

                 newSize = Math.min(newSize, a.length >>> 1);

             @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})

             T[] newArray = (T[]) new Object[newSize];

             tmp = newArray;

         }

         return tmp;

     }

     /**

      * Checks that fromIndex and toIndex are in range, and throws an

      * appropriate exception if they aren't.

      *

      * @param arrayLen the length of the array

      * @param fromIndex the index of the first element of the range

      * @param toIndex the index after the last element of the range

      * @throws IllegalArgumentException if fromIndex > toIndex

      * @throws ArrayIndexOutOfBoundsException if fromIndex < 0

      *         or toIndex > arrayLen

      */

     private static void rangeCheck(int arrayLen, int fromIndex, int toIndex) {

         if (fromIndex > toIndex)

             throw new IllegalArgumentException("fromIndex(" + fromIndex +

                        ") > toIndex(" + toIndex+")");

         if (fromIndex < 0)

             throw new ArrayIndexOutOfBoundsException(fromIndex);

         if (toIndex > arrayLen)

             throw new ArrayIndexOutOfBoundsException(toIndex);

     }

 }