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 
def main():
    from heapq import heapify, heappop, heappushpop,heappush
    n, k, x = map(int, input().split())
    l, sign, h = list(map(int, input().split())), [False] * n, []
    helper = lambda: print(' '.join(str(-a if s else a) for a, s in zip(l, sign)))
    for i, a in enumerate(l):
        if a < 0:
            sign[i] = True
            h.append((-a, i))
        else:
            h.append((a, i))
    heapify(h)
    a, i = heappop(h)
    if 1 - sum(sign) % 2:
        j = min(a // x + (1 if a % x else 0), k)
        a -= j * x
        if a > 0:
            l[i] = a
            return helper()
        k -= j
        a = -a
        sign[i] ^= True
    for _ in range(k):
        a, i = heappushpop(h, (a, i))
        a += x
    l[i] = a
    for a, i in h:
        l[i] = a
    helper()


if __name__ == '__main__':
    main()
