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 main() {
  long long n, k, x, m = 0, i, j, y = 0, f = 0;
  cin >> n >> k >> x;
  long long a[n];
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 0) f++;
    if (a[i] < 0) m++;
    if (abs(a[i]) < abs(a[y])) y = i;
  }
  if (f == 0) {
    if (m % 2 == 0) {
      if (a[y] < 0) {
        j = (-a[y]) / x + 1;
        j = min(k, j);
        a[y] += x * j;
        k -= j;
      } else {
        j = a[y] / x + 1;
        j = min(k, j);
        a[y] -= x * j;
        k -= j;
      }
    }
  } else {
    if (f <= k) {
      bool z = true;
      for (i = 0; i < n; i++) {
        if (a[i] == 0) {
          a[i] += x;
          f--;
          k--;
          if (z && m % 2 == 0) a[i] *= -1, z = false;
        }
      }
    }
  }
  priority_queue<pair<long long, long long>,
                 vector<pair<long long, long long> >,
                 greater<pair<long long, long long> > >
      q;
  for (i = 0; i < n; i++) q.push(pair<long long, long long>(abs(a[i]), i));
  for (j = 0; j < k; j++) {
    i = q.top().second;
    q.pop();
    if (a[i] < 0)
      a[i] -= x;
    else
      a[i] += x;
    q.push(pair<long long, long long>(abs(a[i]), i));
  }
  for (i = 0; i < n; i++) cout << a[i] << (i + 1 == n ? '\n' : ' ');
  return 0;
}
