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;
struct num {
  long long x;
  int id;
  friend bool operator<(const num &a, const num &b) {
    return abs(a.x) > abs(b.x);
  }
} arr[200010];
priority_queue<num> q;
int main() {
  int n, k;
  long long t;
  scanf("%d%d%lld", &n, &k, &t);
  int f = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &arr[i].x);
    if (arr[i].x < 0) {
      f ^= 1;
    }
    q.push((num){arr[i].x, i});
  }
  while (k--) {
    num tt = q.top();
    q.pop();
    if (f) {
      if (tt.x < 0) {
        arr[tt.id].x -= t;
      } else {
        arr[tt.id].x += t;
      }
    } else {
      if (tt.x < 0) {
        arr[tt.id].x += t;
        if (arr[tt.id].x >= 0) {
          f ^= 1;
        }
      } else {
        arr[tt.id].x -= t;
        if (arr[tt.id].x < 0) {
          f ^= 1;
        }
      }
    }
    q.push((num){arr[tt.id].x, tt.id});
  }
  for (int i = 1; i <= n; i++) {
    printf("%lld ", arr[i].x);
  }
  return 0;
}
