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 
#include <bits/stdc++.h>
using namespace std;
int n, k;
long long a[200005], x;
struct heap {
  long long data;
  int indx;
} h[200005];
void heapify(int i) {
  int l, r, s;
  long long tmp;
  l = 2 * i + 1;
  r = 2 * i + 2;
  s = i;
  if (l < n && h[l].data < h[i].data) s = l;
  if (r < n && h[r].data < h[s].data) s = r;
  if (s != i) {
    tmp = h[i].data;
    h[i].data = h[s].data;
    h[s].data = tmp;
    tmp = h[i].indx;
    h[i].indx = h[s].indx;
    h[s].indx = tmp;
    heapify(s);
  }
}
void build() {
  for (int i = 0; i < n; i++) {
    h[i].data = llabs(a[i]);
    h[i].indx = i;
  }
  for (int i = n / 2 - 1; i >= 0; i--) heapify(i);
}
int main() {
  int z = 0, ng = 0;
  scanf("%d%d%I64d", &n, &k, &x);
  for (int i = 0; i < n; i++) scanf("%I64d", &a[i]);
  for (int i = 0; i < n; i++) {
    if (a[i] < 0)
      ng++;
    else if (a[i] == 0)
      z++;
  }
  if (z != 0) {
    if (ng % 2 == 0) {
      for (int i = 0; i < n; i++) {
        if (a[i] == 0) {
          a[i] = -1 * x;
          ng++;
          k--;
          z--;
          break;
        }
      }
    }
    if (k < z) {
      for (int i = 0; i < n; i++) printf("%I64d ", a[i]);
      return 0;
    } else {
      for (int i = 0; i < n; i++) {
        if (a[i] == 0) {
          a[i] = x;
          k--;
          z--;
        }
      }
    }
  }
  build();
  if (ng % 2 == 0) {
    int s = 1, in;
    in = h[0].indx;
    s = s * (a[in] / llabs(a[in]));
    while (k) {
      if (x > h[0].data) {
        s = -1 * s;
        h[0].data = x - h[0].data;
        a[in] = s * h[0].data;
        k--;
        break;
      } else {
        h[0].data = h[0].data - x;
        a[in] = s * h[0].data;
        k--;
      }
    }
  }
  heapify(0);
  while (k) {
    int in = h[0].indx, s;
    s = a[in] / llabs(a[in]);
    h[0].data += x;
    a[in] = s * h[0].data;
    k--;
    heapify(0);
  }
  for (int i = 0; i < n; i++) printf("%I64d ", a[i]);
  return 0;
}
