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;
const int maxm = 2e5 + 10;
struct node {
  long long x;
  int y;
  node(long long xx = 0, int yy = 0) {
    x = xx;
    y = yy;
  }
  bool operator<(const node &head) const { return head.x < x; }
};
priority_queue<node> q;
int m, n;
long long w;
long long s[maxm];
int x, y;
int main() {
  int flag = 0;
  scanf("%d%d", &m, &n);
  cin >> w;
  for (int i = 1; i <= m; i++) {
    scanf("%d", &x);
    s[i] = x;
    if (x < 0) {
      flag ^= 1;
      q.push(node(-x, i));
    } else
      q.push(node(x, i));
  }
  if (flag == 0) {
    while (n) {
      int pos = q.top().y;
      q.pop();
      n--;
      if (s[pos] >= 0) {
        s[pos] -= w;
        q.push(node(abs(s[pos]), pos));
        if (s[pos] < 0) {
          flag ^= 1;
          break;
        }
      } else {
        s[pos] += w;
        q.push(node(abs(s[pos]), pos));
        if (s[pos] >= 0) {
          flag ^= 1;
          break;
        }
      }
    }
  }
  while (n) {
    n--;
    int pos = q.top().y;
    q.pop();
    if (s[pos] >= 0)
      s[pos] += w;
    else
      s[pos] -= w;
    q.push(node(abs(s[pos]), pos));
  }
  for (int i = 1; i <= m; i++) cout << s[i] << " ";
  cout << endl;
  return 0;
}
