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 v[400010];
long long ans[400010];
int main() {
  int n, m, moves;
  scanf(" %d %d %d", &n, &moves, &m);
  int sgn = 1;
  long long min_value = 0;
  int min_pos = -1;
  for (int i = 0; i < n; i++) {
    scanf(" %lld", &v[i]);
    if (v[i] < 0) sgn *= -1;
    if (min_pos == -1 || abs(v[i]) < min_value) {
      min_value = abs(v[i]);
      min_pos = i;
    }
  }
  if (sgn == 1) {
    if (v[min_pos] < 0) {
      while (v[min_pos] <= 0 && moves > 0) {
        v[min_pos] += m;
        moves--;
      }
    } else {
      while (v[min_pos] >= 0 && moves > 0) {
        v[min_pos] -= m;
        moves--;
      }
    }
  }
  for (int i = 0; i < (n); ++i) ans[i] = v[i];
  if (moves > 0) {
    priority_queue<pair<long long, int> > pq;
    for (int i = 0; i < (n); ++i) pq.push(pair<long long, int>(-abs(v[i]), i));
    while (moves > 0) {
      long long v = pq.top().first;
      int ind = pq.top().second;
      pq.pop();
      if (ans[ind] < 0)
        ans[ind] -= m;
      else
        ans[ind] += m;
      pq.push(pair<long long, int>(v - m, ind));
      moves--;
    }
  }
  for (int i = 0; i < (n); ++i) {
    if (i != 0) printf(" ");
    printf("%lld", ans[i]);
  }
  printf("\n");
  return 0;
}
