ComparableTimSortpublic class ComparableTimSort extends Object | This is a near duplicate of {@link TimSort}, modified for use with
arrays of objects that implement {@link Comparable}, instead of using
explicit comparators.
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.) |
| Fields Summary |
|---|
| private static final int | MIN_MERGEThis 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 final Object[] | aThe array being sorted. | | private static final int | MIN_GALLOPWhen we get into galloping mode, we stay there until both runs win less
often than MIN_GALLOP consecutive times. | | private int | minGallopThis 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 static final int | INITIAL_TMP_STORAGE_LENGTHMaximum 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 Object[] | tmpTemp storage for merges. | | private int | stackSizeA 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 final int[] | runBase | | private final int[] | runLen | | private static final boolean | DEBUGAsserts have been placed in if-statements for performace. To enable them,
set this field to true and enable them in VM with a command line flag.
If you modify this class, please do test the asserts! |
| Constructors Summary |
|---|
private ComparableTimSort(Object[] a)Creates a TimSort instance to maintain the state of an ongoing sort.
this.a = a;
// Allocate temp storage (which may be increased later if necessary)
int len = a.length;
@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
Object[] newArray = 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];
|
| Methods Summary |
|---|
| private static void | binarySort(java.lang.Object[] a, int lo, int hi, int start)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.
if (DEBUG) assert lo <= start && start <= hi;
if (start == lo)
start++;
for ( ; start < hi; start++) {
@SuppressWarnings("unchecked")
Comparable<Object> pivot = (Comparable) a[start];
// Set left (and right) to the index where a[start] (pivot) belongs
int left = lo;
int right = start;
if (DEBUG) 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;
}
if (DEBUG) 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 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;
}
| | private static int | countRunAndMakeAscending(java.lang.Object[] a, int lo, int hi)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.
if (DEBUG) 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;
| | private java.lang.Object[] | ensureCapacity(int minCapacity)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.
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"})
Object[] newArray = new Object[newSize];
tmp = newArray;
}
return tmp;
| | private static int | gallopLeft(java.lang.Comparable key, java.lang.Object[] a, int base, int len, int hint)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.
if (DEBUG) 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;
}
if (DEBUG) 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]
}
if (DEBUG) assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs]
return ofs;
| | private static int | gallopRight(java.lang.Comparable key, java.lang.Object[] a, int base, int len, int hint)Like gallopLeft, except that if the range contains an element equal to
key, gallopRight returns the index after the rightmost equal element.
if (DEBUG) 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;
}
if (DEBUG) 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
}
if (DEBUG) assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs]
return ofs;
| | private void | mergeAt(int i)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.
if (DEBUG) assert stackSize >= 2;
if (DEBUG) assert i >= 0;
if (DEBUG) 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];
if (DEBUG) assert len1 > 0 && len2 > 0;
if (DEBUG) 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);
if (DEBUG) 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);
if (DEBUG) 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);
| | private void | mergeCollapse()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.
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
}
}
| | private void | mergeForceCollapse()Merges all runs on the stack until only one remains. This method is
called once, to complete the sort.
while (stackSize > 1) {
int n = stackSize - 2;
if (n > 0 && runLen[n - 1] < runLen[n + 1])
n--;
mergeAt(n);
}
| | private void | mergeHi(int base1, int len1, int base2, int len2)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.)
if (DEBUG) assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
// Copy second run into temp array
Object[] a = this.a; // For performance
Object[] 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;
}
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 {
if (DEBUG) 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 {
if (DEBUG) 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, 0, 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) {
if (DEBUG) 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 {
if (DEBUG) assert len1 == 0;
if (DEBUG) assert len2 > 0;
System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
}
| | private void | mergeLo(int base1, int len1, int base2, int len2)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.)
if (DEBUG) assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
// Copy first run into temp array
Object[] a = this.a; // For performance
Object[] 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;
}
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 {
if (DEBUG) 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 {
if (DEBUG) 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) {
if (DEBUG) 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 {
if (DEBUG) assert len2 == 0;
if (DEBUG) assert len1 > 1;
System.arraycopy(tmp, cursor1, a, dest, len1);
}
| | private static int | minRunLength(int n)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.
if (DEBUG) 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;
| | private void | pushRun(int runBase, int runLen)Pushes the specified run onto the pending-run stack.
this.runBase[stackSize] = runBase;
this.runLen[stackSize] = runLen;
stackSize++;
| | private static void | rangeCheck(int arrayLen, int fromIndex, int toIndex)Checks that fromIndex and toIndex are in range, and throws an
appropriate exception if they aren't.
if (fromIndex > toIndex)
throw new IllegalArgumentException("fromIndex(" + fromIndex +
") > toIndex(" + toIndex+")");
if (fromIndex < 0)
throw new ArrayIndexOutOfBoundsException(fromIndex);
if (toIndex > arrayLen)
throw new ArrayIndexOutOfBoundsException(toIndex);
| | private static void | reverseRange(java.lang.Object[] a, int lo, int hi)Reverse the specified range of the specified array.
hi--;
while (lo < hi) {
Object t = a[lo];
a[lo++] = a[hi];
a[hi--] = t;
}
| | static void | sort(java.lang.Object[] a)
sort(a, 0, a.length);
| | static void | sort(java.lang.Object[] a, int lo, int hi)
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, nRemaining);
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);
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
if (DEBUG) assert lo == hi;
ts.mergeForceCollapse();
if (DEBUG) assert ts.stackSize == 1;
|
|
