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 = [int(val) for val in input().split()]
a = [int(val) for val in input().split()]

data = [(val, 1, i) if val >= 0 else (-1 * val, -1, i) for i, val in enumerate(a)]
heapq.heapify(data)
sign = sum([1 if s == -1 else 0 for _, s, _ in data])

if sign % 2 == 1:
    for i in range(k):
        e = heapq.heappop(data)
        heapq.heappush(data, (e[0] + x, e[1], e[2]))
else:
    e = heapq.heappop(data)
    if e[0] <  k * x:
        s = (e[0] // x) + 1
        k -= s
        heapq.heappush(data, (s * x - e[0], -1 * e[1], e[2]))
        for i in range(k):
            e = heapq.heappop(data)
            heapq.heappush(data, (e[0] + x, e[1], e[2]))
    else:
        heapq.heappush(data, (e[0] - k * x, e[1], e[2]))

output = [0] * n
for val, s, i in data:
    output[i] = s * val
print(' '.join([str(val) for val in output]))
   
