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	/*
	* Copyright (c) 2009, 2013, Oracle and/or its affiliates. All rights reserved.
	* 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
	* or visit www.oracle.com if you need additional information or have any
	* questions.
	*/

	package java.util;

	/**
	* This is a near duplicate of {@link TimSort}, modified for use with
	* arrays of objects that implement {@link Comparable}, instead of using
	* explicit comparators.
	*
	* <p>If you are using an optimizing VM, you may find that ComparableTimSort
	* offers no performance benefit over TimSort in conjunction with a
	* comparator that simply returns {@code ((Comparable)first).compareTo(Second)}.
	* If this is the case, you are better off deleting ComparableTimSort to
	* eliminate the code duplication. (See Arrays.java for details.)
	*
	* @author Josh Bloch
	*/
	class ComparableTimSort {
	/**
	* 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 Object[] a;

	/**
	* 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. A workspace array may optionally be
	* provided in constructor, and if so will be used as long as it
	* is big enough.
	*/
	private Object[] tmp;
	private int tmpBase; // base of tmp array slice
	private int tmpLen; // length of tmp array slice

	/**
	* 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 work a workspace array (slice)
	* @param workBase origin of usable space in work array
	* @param workLen usable size of work array
	*/
	private ComparableTimSort(Object[] a, Object[] work, int workBase, int workLen) {
	this.a = a;

	// Allocate temp storage (which may be increased later if necessary)
	int len = a.length;
	int tlen = (len < 2 * INITIAL_TMP_STORAGE_LENGTH) ?
	len >>> 1 : INITIAL_TMP_STORAGE_LENGTH;
	if (work == null \|\| workLen < tlen \|\| workBase + tlen > work.length) {
	tmp = new Object[tlen];
	tmpBase = 0;
	tmpLen = tlen;
	}
	else {
	tmp = work;
	tmpBase = workBase;
	tmpLen = workLen;
	}

	/*
	* 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.
	* The maximum value of 49 allows for an array up to length
	* Integer.MAX_VALUE-4, if array is filled by the worst case stack size
	* increasing scenario. More explanations are given in section 4 of:
	* http://envisage-project.eu/wp-content/uploads/2015/02/sorting.pdf
	*/
	int stackLen = (len < 120 ? 5 :
	len < 1542 ? 10 :
	len < 119151 ? 24 : 49);
	runBase = new int[stackLen];
	runLen = new int[stackLen];
	}

	/*
	* The next method (package private and static) constitutes the
	* entire API of this class.
	*/

	/**
	* Sorts the given range, using the given workspace array slice
	* for temp storage when possible. This method is designed to be
	* invoked from public methods (in class Arrays) after performing
	* any necessary array bounds checks and expanding parameters into
	* the required forms.
	*
	* @param a the array to be sorted
	* @param lo the index of the first element, inclusive, to be sorted
	* @param hi the index of the last element, exclusive, to be sorted
	* @param work a workspace array (slice)
	* @param workBase origin of usable space in work array
	* @param workLen usable size of work array
	* @since 1.8
	*/
	static void sort(Object[] a, int lo, int hi, Object[] work, int workBase, int workLen) {
	assert a != null && lo >= 0 && lo <= hi && hi <= a.length;

	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);
	binarySort(a, lo, hi, lo + initRunLen);
	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.
	*/
	ComparableTimSort ts = new ComparableTimSort(a, work, workBase, workLen);
	int minRun = minRunLength(nRemaining);
	do {
	// Identify next run
	int runLen = countRunAndMakeAscending(a, lo, hi);

	// 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);
	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})
	*/
	@SuppressWarnings({"fallthrough", "rawtypes", "unchecked"})
	private static void binarySort(Object[] a, int lo, int hi, int start) {
	assert lo <= start && start <= hi;
	if (start == lo)
	start++;
	for ( ; start < hi; start++) {
	Comparable pivot = (Comparable) 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 (pivot.compareTo(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}.
	* @return the length of the run beginning at the specified position in
	* the specified array
	*/
	@SuppressWarnings({"unchecked", "rawtypes"})
	private static int countRunAndMakeAscending(Object[] a, int lo, int hi) {
	assert lo < hi;
	int runHi = lo + 1;
	if (runHi == hi)
	return 1;

	// Find end of run, and reverse range if descending
	if (((Comparable) a[runHi++]).compareTo(a[lo]) < 0) { // Descending
	while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) < 0)
	runHi++;
	reverseRange(a, lo, runHi);
	} else { // Ascending
	while (runHi < hi && ((Comparable) a[runHi]).compareTo(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
	*/
	@SuppressWarnings("unchecked")
	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((Comparable<Object>) a[base2], a, base1, len1, 0);
	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((Comparable<Object>) a[base1 + len1 - 1], a,
	base2, len2, len2 - 1);
	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.
	* @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 int gallopLeft(Comparable<Object> key, Object[] a,
	int base, int len, int hint) {
	assert len > 0 && hint >= 0 && hint < len;

	int lastOfs = 0;
	int ofs = 1;
	if (key.compareTo(a[base + hint]) > 0) {
	// Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
	int maxOfs = len - hint;
	while (ofs < maxOfs && key.compareTo(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 && key.compareTo(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 (key.compareTo(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.
	* @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
	*/
	private static int gallopRight(Comparable<Object> key, Object[] a,
	int base, int len, int hint) {
	assert len > 0 && hint >= 0 && hint < len;

	int ofs = 1;
	int lastOfs = 0;
	if (key.compareTo(a[base + hint]) < 0) {
	// Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
	int maxOfs = hint + 1;
	while (ofs < maxOfs && key.compareTo(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 && key.compareTo(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 (key.compareTo(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)
	*/
	@SuppressWarnings({"unchecked", "rawtypes"})
	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
	Object[] a = this.a; // For performance
	Object[] tmp = ensureCapacity(len1);

	int cursor1 = tmpBase; // Indexes into tmp array
	int cursor2 = base2; // Indexes int a
	int dest = base1; // Indexes int a
	System.arraycopy(a, base1, tmp, cursor1, len1);

	// 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;
	}

	int minGallop = this.minGallop; // Use local variable for performance
	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 (((Comparable) a[cursor2]).compareTo(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((Comparable) a[cursor2], tmp, cursor1, len1, 0);
	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((Comparable) tmp[cursor1], a, cursor2, len2, 0);
	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)
	*/
	@SuppressWarnings({"unchecked", "rawtypes"})
	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
	Object[] a = this.a; // For performance
	Object[] tmp = ensureCapacity(len2);
	int tmpBase = this.tmpBase;
	System.arraycopy(a, base2, tmp, tmpBase, len2);

	int cursor1 = base1 + len1 - 1; // Indexes into a
	int cursor2 = tmpBase + 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, tmpBase, 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;
	}

	int minGallop = this.minGallop; // Use local variable for performance
	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 (((Comparable) tmp[cursor2]).compareTo(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((Comparable) tmp[cursor2], a, base1, len1, len1 - 1);
	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((Comparable) a[cursor1], tmp, tmpBase, len2, len2 - 1);
	if (count2 != 0) {
	dest -= count2;
	cursor2 -= count2;
	len2 -= count2;
	System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
	if (len2 <= 1)
	break outer; // len2 == 1 \|\| len2 == 0
	}
	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, tmpBase, 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 Object[] ensureCapacity(int minCapacity) {
	if (tmpLen < 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"})
	Object[] newArray = new Object[newSize];
	tmp = newArray;
	tmpLen = newSize;
	tmpBase = 0;
	}
	return tmp;
	}

	}

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