anchor.diff:  	var/list/runLens = list()
   
   
  -	proc/timSort(start, end)
  -		runBases.Cut()
  -		runLens.Cut()
  -
  -		var/remaining = end - start
  -
  -		//If array is small, do a 'mini-TimSort' with no merges
  -		if(remaining < MIN_MERGE)
  -			var/initRunLen = countRunAndMakeAscending(start, end)
  -			binarySort(start, end, start+initRunLen)
  -			return
  -
  -		//March over the array finding natural runs
  -		//Extend any short natural runs to runs of length minRun
  -		var/minRun = minRunLength(remaining)
  -
  -		do
  -				//identify next run
  -			var/runLen = countRunAndMakeAscending(start, end)
  -
  -				//if run is short, extend to min(minRun, remaining)
  -			if(runLen < minRun)
  -				var/force = (remaining <= minRun) ? remaining : minRun
  -
  -				binarySort(start, start+force, start+runLen)
  -				runLen = force
  -
  -				//add data about run to queue
  -			runBases.Add(start)
  -			runLens.Add(runLen)
  -
  -				//maybe merge
  -			mergeCollapse()
  -
  -				//Advance to find next run
  -			start += runLen
  -			remaining -= runLen
  -
  -		while(remaining > 0)
  -
  -
  -			//Merge all remaining runs to complete sort
  -		//ASSERT(start == end)
  -		mergeForceCollapse();
  -		//ASSERT(runBases.len == 1)
  -
  -			//reset minGallop, for successive calls
  -		minGallop = MIN_GALLOP
  -
  -		return L
  -
  -	/*
  +/datum/sortInstance/proc/timSort(start, end)
  +	runBases.Cut()
  +	runLens.Cut()
  +
  +	var/remaining = end - start
  +
  +	//If array is small, do a 'mini-TimSort' with no merges
  +	if(remaining < MIN_MERGE)
  +		var/initRunLen = countRunAndMakeAscending(start, end)
  +		binarySort(start, end, start+initRunLen)
  +		return
  +
  +	//March over the array finding natural runs
  +	//Extend any short natural runs to runs of length minRun
  +	var/minRun = minRunLength(remaining)
  +
  +	do
  +			//identify next run
  +		var/runLen = countRunAndMakeAscending(start, end)
  +
  +			//if run is short, extend to min(minRun, remaining)
  +		if(runLen < minRun)
  +			var/force = (remaining <= minRun) ? remaining : minRun
  +
  +			binarySort(start, start+force, start+runLen)
  +			runLen = force
  +
  +			//add data about run to queue
  +		runBases.Add(start)
  +		runLens.Add(runLen)
  +
  +			//maybe merge
  +		mergeCollapse()
  +
  +			//Advance to find next run
  +		start += runLen
  +		remaining -= runLen
  +
  +	while(remaining > 0)
  +
  +
  +		//Merge all remaining runs to complete sort
  +	//ASSERT(start == end)
  +	mergeForceCollapse();
  +	//ASSERT(runBases.len == 1)
  +
  +		//reset minGallop, for successive calls
  +	minGallop = MIN_GALLOP
  +
  +	return L
  +
  +/*
   	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
  
positive.code: 	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.
  	*/
  	proc/countRunAndMakeAscending(lo, hi)
  		//ASSERT(lo < hi)
  
  		var/runHi = lo + 1
  		if(runHi >= hi)
  			return 1
  
  		var/last = fetchElement(L,lo)
  		var/current = fetchElement(L,runHi++)
  
  		if(call(cmp)(current, last) < 0)
  			while(runHi < hi)
  				last = current
  				current = fetchElement(L,runHi)
  				if(call(cmp)(current, last) >= 0)
  					break
  				++runHi
  			reverseRange(L, lo, runHi)
  		else
  			while(runHi < hi)
  				last = current
  				current = fetchElement(L,runHi)
  				if(call(cmp)(current, last) < 0)
  					break
  				++runHi
  
  		return runHi - lo
  
  	//Returns the minimum acceptable run length for an array of the specified length.
  	//Natural runs shorter than this will be extended with binarySort
  	proc/minRunLength(n)
  		//ASSERT(n >= 0)
  		var/r = 0	//becomes 1 if any bits are shifted off
  		while(n >= MIN_MERGE)
  			r |= (n & 1)
  			n >>= 1
  		return n + r
  
  	//Examines the stack of runs waiting to be merged and merges adjacent runs until the stack invariants are reestablished:
  	//	runLen[i-3] > runLen[i-2] + runLen[i-1]
  	//	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
  	proc/mergeCollapse()
  		while(runBases.len >= 2)
  			var/n = runBases.len - 1
  			if(n > 1 && runLens[n-1] <= runLens[n] + runLens[n+1])
  				if(runLens[n-1] < runLens[n+1])
  					--n
  				mergeAt(n)
  			else if(runLens[n] <= runLens[n+1])
  				mergeAt(n)
  			else
  				break	//Invariant is established
  
  
  	//Merges all runs on the stack until only one remains.
  	//Called only once, to finalise the sort
  	proc/mergeForceCollapse()
  		while(runBases.len >= 2)
  			var/n = runBases.len - 1
  			if(n > 1 && runLens[n-1] < runLens[n+1])
  				--n
  			mergeAt(n)
  
  
  	//Merges the two consecutive 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
  	proc/mergeAt(i)
  		//ASSERT(runBases.len >= 2)
  		//ASSERT(i >= 1)
  		//ASSERT(i == runBases.len - 1 || i == runBases.len - 2)
  
  		var/base1 = runBases[i]
  		var/base2 = runBases[i+1]
  		var/len1 = runLens[i]
  		var/len2 = runLens[i+1]
  
  		//ASSERT(len1 > 0 && len2 > 0)
  		//ASSERT(base1 + len1 == base2)
  
  		//Record the legth 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.
  		runLens[i] += runLens[i+1]
  		runLens.Cut(i+1, i+2)
  		runBases.Cut(i+1, i+2)
  
  
  		//Find where the first element of run2 goes in run1.
  		//Prior elements in run1 can be ignored (because they're already in place)
  		var/k = gallopRight(fetchElement(L,base2), 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(fetchElement(L,base1 + len1 - 1), 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 to insert key within the specified sorted range
  		If the range contains elements equal to key, this will return the index of the LEFTMOST of those elements
  
  		key		the element to be inserted into the sorted range
  		base	the index of the first element of the sorted range
  		len		the length of the sorted range, must be greater than 0
  		hint	the offset from base at which to begin the search, such that 0 <= hint < len; i.e. base <= hint < base+hint
  
  		Returns the index at which to insert element 'key'
  	*/
  	proc/gallopLeft(key, base, len, hint)
  		//ASSERT(len > 0 && hint >= 0 && hint < len)
  
  		var/lastOffset = 0
  		var/offset = 1
  		if(call(cmp)(key, fetchElement(L,base+hint)) > 0)
  			var/maxOffset = len - hint
  			while(offset < maxOffset && call(cmp)(key, fetchElement(L,base+hint+offset)) > 0)
  				lastOffset = offset
  				offset = (offset << 1) + 1
  
  			if(offset > maxOffset)
  				offset = maxOffset
  
  			lastOffset += hint
  			offset += hint
  
  		else
  			var/maxOffset = hint + 1
  			while(offset < maxOffset && call(cmp)(key, fetchElement(L,base+hint-offset)) <= 0)
  				lastOffset = offset
  				offset = (offset << 1) + 1
  
  			if(offset > maxOffset)
  				offset = maxOffset
  
  			var/temp = lastOffset
  			lastOffset = hint - offset
  			offset = hint - temp
  
  			//ASSERT(-1 <= lastOffset && lastOffset < offset && offset <= len)
  
  		//Now L[base+lastOffset] < key <= L[base+offset], so key belongs somewhere to the right of lastOffset but no farther than
  		//offset. Do a binary search with invariant L[base+lastOffset-1] < key <= L[base+offset]
  		++lastOffset
  		while(lastOffset < offset)
  			var/m = lastOffset + ((offset - lastOffset) >> 1)
  
  			if(call(cmp)(key, fetchElement(L,base+m)) > 0)
  				lastOffset = m + 1
  			else
  				offset = m
  
  		//ASSERT(lastOffset == offset)
  		return offset
  
  	/**
  	 * 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]
  	 */
  	proc/gallopRight(key, base, len, hint)
  		//ASSERT(len > 0 && hint >= 0 && hint < len)
  
  		var/offset = 1
  		var/lastOffset = 0
  		if(call(cmp)(key, fetchElement(L,base+hint)) < 0)	//key <= L[base+hint]
  			var/maxOffset = hint + 1	//therefore we want to insert somewhere in the range [base,base+hint] = [base+,base+(hint+1))
  			while(offset < maxOffset && call(cmp)(key, fetchElement(L,base+hint-offset)) < 0)	//we are iterating backwards
  				lastOffset = offset
  				offset = (offset << 1) + 1	//1 3 7 15
  
  			if(offset > maxOffset)
  				offset = maxOffset
  
  			var/temp = lastOffset
  			lastOffset = hint - offset
  			offset = hint - temp
  
  		else	//key > L[base+hint]
  			var/maxOffset = len - hint	//therefore we want to insert somewhere in the range (base+hint,base+len) = [base+hint+1, base+hint+(len-hint))
  			while(offset < maxOffset && call(cmp)(key, fetchElement(L,base+hint+offset)) >= 0)
  				lastOffset = offset
  				offset = (offset << 1) + 1
  
  			if(offset > maxOffset)
  				offset = maxOffset
  
  			lastOffset += hint
  			offset += hint
  
  		//ASSERT(-1 <= lastOffset && lastOffset < offset && offset <= len)
  
  		++lastOffset
  		while(lastOffset < offset)
  			var/m = lastOffset + ((offset - lastOffset) >> 1)
  
  			if(call(cmp)(key, fetchElement(L,base+m)) < 0)	//key <= L[base+m]
  				offset = m
  			else							//key > L[base+m]
  				lastOffset = m + 1
  
  		//ASSERT(lastOffset == offset)
  
  		return offset
  
  
  	//Merges two adjacent runs in-place in a stable fashion.
  	//For performance this method should only be called when len1 <= len2!
  	proc/mergeLo(base1, len1, base2, len2)
  		//ASSERT(len1 > 0 && len2 > 0 && base1 + len1 == base2)
  
  		var/cursor1 = base1
  		var/cursor2 = base2
  
  		//degenerate cases
  		if(len2 == 1)
  			moveElement(L, cursor2, cursor1)
  			return
  
  		if(len1 == 1)
  			moveElement(L, cursor1, cursor2+len2)
  			return
  
  
  		//Move first element of second run
  		moveElement(L, cursor2++, cursor1++)
  		--len2
  
  		outer:
  			while(1)
  				var/count1 = 0	//# of times in a row that first run won
  				var/count2 = 0	//	"	"	"	"	"	"  second run won
  
  				//do the straightfoward thin until one run starts winning consistently
  
  				do
  					//ASSERT(len1 > 1 && len2 > 0)
  					if(call(cmp)(fetchElement(L,cursor2), fetchElement(L,cursor1)) < 0)
  						moveElement(L, cursor2++, cursor1++)
  						--len2
  
  						++count2
  						count1 = 0
  
  						if(len2 == 0)
  							break outer
  					else
  						++cursor1
  
  						++count1
  						count2 = 0
  
  						if(--len1 == 1)
  							break outer
  
  				while((count1 | count2) < minGallop)
  
  
  				//one run is winning consistently so galloping may provide huge benifits
  				//so try galloping, until such time as the run is no longer consistently winning
  				do
  					//ASSERT(len1 > 1 && len2 > 0)
  
  					count1 = gallopRight(fetchElement(L,cursor2), cursor1, len1, 0)
  					if(count1)
  						cursor1 += count1
  						len1 -= count1
  
  						if(len1 <= 1)
  							break outer
  
  					moveElement(L, cursor2, cursor1)
  					++cursor2
  					++cursor1
  					if(--len2 == 0)
  						break outer
  
  					count2 = gallopLeft(fetchElement(L,cursor1), cursor2, len2, 0)
  					if(count2)
  						moveRange(L, cursor2, cursor1, count2)
  
  						cursor2 += count2
  						cursor1 += count2
  						len2 -= count2
  
  						if(len2 == 0)
  							break outer
  
  					++cursor1
  					if(--len1 == 1)
  						break outer
  
  					--minGallop
  
  				while((count1|count2) > MIN_GALLOP)
  
  				if(minGallop < 0)
  					minGallop = 0
  				minGallop += 2;  // Penalize for leaving gallop mode
  
  
  		if(len1 == 1)
  			//ASSERT(len2 > 0)
  			moveElement(L, cursor1, cursor2+len2)
  
  		//else
  			//ASSERT(len2 == 0)
  			//ASSERT(len1 > 1)
  
  
  	proc/mergeHi(base1, len1, base2, len2)
  		//ASSERT(len1 > 0 && len2 > 0 && base1 + len1 == base2)
  
  		var/cursor1 = base1 + len1 - 1	//start at end of sublists
  		var/cursor2 = base2 + len2 - 1
  
  		//degenerate cases
  		if(len2 == 1)
  			moveElement(L, base2, base1)
  			return
  
  		if(len1 == 1)
  			moveElement(L, base1, cursor2+1)
  			return
  
  		moveElement(L, cursor1--, cursor2-- + 1)
  		--len1
  
  		outer:
  			while(1)
  				var/count1 = 0	//# of times in a row that first run won
  				var/count2 = 0	//	"	"	"	"	"	"  second run won
  
  				//do the straightfoward thing until one run starts winning consistently
  				do
  					//ASSERT(len1 > 0 && len2 > 1)
  					if(call(cmp)(fetchElement(L,cursor2), fetchElement(L,cursor1)) < 0)
  						moveElement(L, cursor1--, cursor2-- + 1)
  						--len1
  
  						++count1
  						count2 = 0
  
  						if(len1 == 0)
  							break outer
  					else
  						--cursor2
  						--len2
  
  						++count2
  						count1 = 0
  
  						if(len2 == 1)
  							break outer
  				while((count1 | count2) < minGallop)
  
  				//one run is winning consistently so galloping may provide huge benifits
  				//so try galloping, until such time as the run is no longer consistently winning
  				do
  					//ASSERT(len1 > 0 && len2 > 1)
  
  					count1 = len1 - gallopRight(fetchElement(L,cursor2), base1, len1, len1-1)	//should cursor1 be base1?
  					if(count1)
  						cursor1 -= count1
  
  						moveRange(L, cursor1+1, cursor2+1, count1)	//cursor1+1 == cursor2 by definition
  
  						cursor2 -= count1
  						len1 -= count1
  
  						if(len1 == 0)
  							break outer
  
  					--cursor2
  
  					if(--len2 == 1)
  						break outer
  
  					count2 = len2 - gallopLeft(fetchElement(L,cursor1), cursor1+1, len2, len2-1)
  					if(count2)
  						cursor2 -= count2
  						len2 -= count2
  
  						if(len2 <= 1)
  							break outer
  
  					moveElement(L, cursor1--, cursor2-- + 1)
  					--len1
  
  					if(len1 == 0)
  						break outer
  
  					--minGallop
  				while((count1|count2) > MIN_GALLOP)
  
  				if(minGallop < 0)
  					minGallop = 0
  				minGallop += 2	// Penalize for leaving gallop mode
  
  		if(len2 == 1)
  			//ASSERT(len1 > 0)
  
  			cursor1 -= len1
  			moveRange(L, cursor1+1, cursor2+1, len1)
  
  		//else
  			//ASSERT(len1 == 0)
  			//ASSERT(len2 > 0)
  
  
  	proc/mergeSort(start, end)
  		var/remaining = end - start
  
  		//If array is small, do an insertion sort
  		if(remaining < MIN_MERGE)
  			binarySort(start, end, start/*+initRunLen*/)
  			return
  
  		var/minRun = minRunLength(remaining)
  
  		do
  			var/runLen = (remaining <= minRun) ? remaining : minRun
  
  			binarySort(start, start+runLen, start)
  
  			//add data about run to queue
  			runBases.Add(start)
  			runLens.Add(runLen)
  
  			//Advance to find next run
  			start += runLen
  			remaining -= runLen
  
  		while(remaining > 0)
  
  		while(runBases.len >= 2)
  			var/n = runBases.len - 1
  			if(n > 1 && runLens[n-1] <= runLens[n] + runLens[n+1])
  				if(runLens[n-1] < runLens[n+1])
  					--n
  				mergeAt2(n)
  			else if(runLens[n] <= runLens[n+1])
  				mergeAt2(n)
  			else
  				break	//Invariant is established
  
  		while(runBases.len >= 2)
  			var/n = runBases.len - 1
  			if(n > 1 && runLens[n-1] < runLens[n+1])
  				--n
  			mergeAt2(n)
  
  		return L
  
  	proc/mergeAt2(i)
  		var/cursor1 = runBases[i]
  		var/cursor2 = runBases[i+1]
  
  		var/end1 = cursor1+runLens[i]
  		var/end2 = cursor2+runLens[i+1]
  
  		var/val1 = fetchElement(L,cursor1)
  		var/val2 = fetchElement(L,cursor2)
  
  		while(1)
  			if(call(cmp)(val1,val2) <= 0)
  				if(++cursor1 >= end1)
  					break
  				val1 = fetchElement(L,cursor1)
  			else
  				moveElement(L,cursor2,cursor1)
  
  				if(++cursor2 >= end2)
  					break
  				++end1
  				++cursor1
  
  				val2 = fetchElement(L,cursor2)
  
  
  		//Record the legth 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.
  		runLens[i] += runLens[i+1]
  		runLens.Cut(i+1, i+2)
  		runBases.Cut(i+1, i+2)
  
  #undef MIN_GALLOP
  #undef MIN_MERGE
  
negative.code: 	lo		the index of the first element in the range to be sorted
  	hi		the index after the last element in the range to be sorted
  	start	the index of the first element in the range that is	not already known to be sorted
  */
  /datum/sortInstance/proc/binarySort(lo, hi, start)
  	//ASSERT(lo <= start && start <= hi)
  	if(start <= lo)
  		start = lo + 1
  
  	for(,start < hi, ++start)
  		var/pivot = fetchElement(L,start)
  
  		//set left and right to the index where pivot belongs
  		var/left = lo
  		var/right = start
  		//ASSERT(left <= right)
  
  		//[lo, left) elements <= pivot < [right, start) elements
  		//in other words, find where the pivot element should go using bisection search
  		while(left < right)
  			var/mid = (left + right) >> 1	//round((left+right)/2)
  			if(call(cmp)(fetchElement(L,mid), pivot) > 0)
  				right = mid
  			else
  				left = mid+1
  
  		//ASSERT(left == right)
  		moveElement(L, start, left)	//move pivot element to correct location in the sorted range
  
  /*
  	Returns the length of the run beginning at the specified position and reverses the run if it is back-to-front
  
  	A run is the longest ascending sequence with:
  