Recently Maxim has found an array of n integers, needed by no one. He immediately come up with idea of changing it: he invented positive integer x and decided to add or subtract it from arbitrary array elements. Formally, by applying single operation Maxim chooses integer i (1 ≤ i ≤ n) and replaces the i-th element of array ai either with ai + x or with ai - x. Please note that the operation may be applied more than once to the same position.

Maxim is a curious minimalis, thus he wants to know what is the minimum value that the product of all array elements (i.e. <image>) can reach, if Maxim would apply no more than k operations to it. Please help him in that.

Input

The first line of the input contains three integers n, k and x (1 ≤ n, k ≤ 200 000, 1 ≤ x ≤ 109) — the number of elements in the array, the maximum number of operations and the number invented by Maxim, respectively.

The second line contains n integers a1, a2, ..., an (<image>) — the elements of the array found by Maxim.

Output

Print n integers b1, b2, ..., bn in the only line — the array elements after applying no more than k operations to the array. In particular, <image> should stay true for every 1 ≤ i ≤ n, but the product of all array elements should be minimum possible.

If there are multiple answers, print any of them.

Examples

Input

5 3 1
5 4 3 5 2


Output

5 4 3 5 -1 


Input

5 3 1
5 4 3 5 5


Output

5 4 0 5 5 


Input

5 3 1
5 4 4 5 5


Output

5 1 4 5 5 


Input

3 2 7
5 4 2


Output

5 11 -5 
import heapq
s = input().split()
n = int(s[0])
k = int(s[1])
x = int(s[2])
neg=0
zeroes=0
numbers = list(map(int,input().split()))
num = []
for i in numbers:
	if(i<0):
		neg+=1
	elif(i==0):
		zeroes+=1

for i,x1 in enumerate(numbers):
	num.append([abs(x1),i])
heapq.heapify(num)
temp=[]
steps=0
if(neg%2==0):
	temp = heapq.heappop(num)
	# print(temp[1],temp[0])
	# print(numbers[temp[1]],x)
	if(numbers[temp[1]]<0):
		steps = min(abs(numbers[temp[1]])//x+1,k)
		k-=steps
		numbers[temp[1]]+=steps*x
		heapq.heappush(num,[abs(numbers[temp[1]]),temp[1]])
	else:
		steps = min((abs(numbers[temp[1]])//x)+1,k)
		# print(steps)
		k-=steps
		numbers[temp[1]]-=steps*x
		heapq.heappush(num,[abs(numbers[temp[1]]),temp[1]])

while(k>0):
	temp = heapq.heappop(num)
	# print(temp[1],temp[0])
	if(numbers[temp[1]]<0):
		# steps = (numbers[temp[1]]//x+1)
		# k-=steps
		numbers[temp[1]]-=x
		heapq.heappush(num,[abs(numbers[temp[1]]),temp[1]])
	else:
		# steps = (numbers[temp[1]]//x+1)
		# k-=steps
		numbers[temp[1]]+=x
		heapq.heappush(num,[abs(numbers[temp[1]]),temp[1]])
	k-=1
for i in numbers:
	print(i,end=" ")

