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;
long long f[222222] = {};
long long n, m, k;
long long cnt = 0, pos, minn = 2147483647;
priority_queue<pair<long long, long long> > q;
int main() {
  scanf("%I64d%I64d%I64d", &n, &m, &k);
  for (int i = 1; i <= n; i++) scanf("%I64d", &f[i]);
  for (int i = 1; i <= n; i++) {
    if (f[i] < 0) cnt++;
    if (abs(f[i]) < minn) minn = abs(f[i]), pos = i;
  }
  if (!(cnt & 1)) {
    if (m * k <= minn) {
      if (f[pos] < 0)
        f[pos] += m * k;
      else
        f[pos] -= m * k;
      for (int i = 1; i <= n; i++) printf("%d ", f[i]);
      return 0;
    } else {
      long long temp = abs(f[pos]);
      temp = temp / k + 1;
      m -= temp;
      temp *= k;
      if (f[pos] < 0)
        f[pos] += temp;
      else
        f[pos] -= temp;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (f[i] < 0)
      q.push(make_pair(f[i] * 2 - 1, i));
    else
      q.push(make_pair(-f[i] * 2, i));
  }
  while (m) {
    m--;
    long long temp1 = q.top().first;
    long long temp2 = q.top().second;
    q.pop();
    temp1 -= 2 * k;
    q.push(make_pair(temp1, temp2));
  }
  while (!q.empty()) {
    long long temp1 = q.top().first;
    long long temp2 = q.top().second;
    temp1 = -temp1;
    if (temp1 & 1)
      f[temp2] = -temp1 / 2;
    else
      f[temp2] = temp1 / 2;
    q.pop();
  }
  for (int i = 1; i <= n; i++) printf("%I64d\n", f[i]);
}
