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 n, k, x;
struct node {
  long long num;
  int id;
  node(long long num, int id) : num(num), id(id) {}
  bool operator<(const node& b) const { return num > b.num; }
};
priority_queue<node> pq;
long long a[200010];
int main() {
  while (scanf("%I64d %I64d %I64d", &n, &k, &x) == 3) {
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%I64d", &a[i]);
      if (a[i] < 0) cnt++;
      pq.push(node(abs(a[i]), i));
    }
    while (k--) {
      int id = pq.top().id;
      pq.pop();
      if (cnt % 2) {
        if (a[id] < 0) {
          a[id] -= x;
        } else {
          a[id] += x;
        }
      } else {
        if (a[id] < 0) {
          a[id] += x;
          if (a[id] >= 0) cnt--;
        } else {
          a[id] -= x;
          if (a[id] < 0) cnt++;
        }
      }
      pq.push(node(abs(a[id]), id));
    }
    for (int i = 1; i <= n; i++) {
      if (i != 1) printf(" ");
      printf("%I64d", a[i]);
    }
    puts("");
  }
}
