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

n, k, X = map(int, input().split())
a       = list(map(int, input().split()))
Q       = []
sit     = 0

for i, x in enumerate(a):
    if x < 0:
        sit += 1
    heapq.heappush(Q, (abs(x), x, i))    
        
sit = sit % 2

while k > 0:
    (val, x, i) = heapq.heappop(Q)
    change = None
    
    if x < 0:
        if sit == 1:
            change = x - X
        else:
            change = x + X
    else:
        if sit == 1:
            change = x + X
        else:
            change = x - X
    
    if (x * change < 0) or (x < 0 and change == 0) or (x == 0 and change < 0):
        sit = (sit+1) % 2
        
    heapq.heappush(Q, (abs(change), change, i))
    k-=1
    
ans = [0] * n
for (val, x, i) in Q:
    ans[i] = str(x)
print(' '.join(ans))    

#5 3 1
#5 4 3 5 2

#5 3 1
#5 4 3 5 5

#5 3 1
#5 4 4 5 5

#3 2 7
#5 4 2