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 abss(long long x) { return x > 0 ? x : -x; }
struct fff {
  long long num;
  int p;
  bool operator<(const fff &a) const { return abss(num) > abss(a.num); }
};
fff f[200001];
priority_queue<fff> q;
int main() {
  int n, k, x;
  scanf("%d%d%d", &n, &k, &x);
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &f[i].num);
    if (f[i].num < 0) cnt++;
    f[i].p = i + 1;
    q.push(f[i]);
  }
  while (k) {
    fff now = q.top();
    q.pop();
    if (now.num) {
      q.push(now);
      break;
    }
    if (cnt % 2 == 0) {
      now.num = now.num - x;
      cnt++;
    } else
      now.num = now.num + x;
    q.push(now);
    k--;
  }
  if (k) {
    if (cnt % 2 == 0) {
      fff now = q.top();
      q.pop();
      if (now.num > 0) {
        while (now.num >= 0 && k) {
          now.num = now.num - x;
          k--;
        }
      } else {
        while (now.num <= 0 && k) {
          now.num = now.num + x;
          k--;
        }
      }
      q.push(now);
    }
    while (k) {
      fff now = q.top();
      q.pop();
      if (now.num > 0)
        now.num = now.num + x;
      else
        now.num = now.num - x;
      q.push(now);
      k--;
    }
  }
  while (!q.empty()) {
    fff now = q.top();
    q.pop();
    f[now.p - 1] = now;
  }
  for (int i = 0; i < n; i++) {
    if (i) printf(" ");
    printf("%I64d", f[i].num);
  }
  printf("\n");
  return 0;
}
