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

inp = list(map(int, input().split()))
n = inp[0]
k = inp[1]
x = inp[2]

a = [int(i) for i in input().strip().split(' ')]

isPositive = True
nodes = [[abs(val), val, index] for index, val in enumerate(a)]

for el in a:
    if el < 0:
        isPositive = not isPositive

heapq.heapify(nodes)

for i in range(k):
    minNode = nodes[0]
    val = minNode[1]
    isCurPositive = val >= 0
    newVal = val
    if isPositive == isCurPositive:
        newVal -= x
    else:
        newVal += x

    minNode[1] = newVal
    if val >= 0 > newVal or val < 0 <= newVal:
        isPositive = not isPositive

    heapq.heapreplace(nodes, [abs(newVal), newVal, minNode[2]])

result = [None] * n
for node in nodes:
    result[node[2]] = str(node[1])

print(" ".join(result))
