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 ans[200005];
long long ask[200005];
typedef struct node {
  long long x;
  long long ab;
  int id;
  bool operator<(const node &cx) const { return cx.ab < ab; }
} ss;
priority_queue<ss> que;
int main(void) {
  long long n, k, x;
  while (scanf("%lld %lld %lld", &n, &k, &x) != EOF) {
    int fu = 0;
    for (int i = 0; i < n; i++) {
      scanf("%lld", &ans[i]);
      if (ans[i] < 0) {
        fu++;
      }
      ss ak;
      ak.ab = abs(ans[i]);
      ak.x = ans[i];
      ak.id = i;
      que.push(ak);
    }
    while (k) {
      k--;
      ss head = que.top();
      que.pop();
      if (head.x < 0) {
        fu--;
        if (fu % 2) {
          head.x += x;
          head.ab = abs(head.x);
          que.push(head);
          if (head.x < 0) fu++;
        } else {
          head.x -= x;
          head.ab = abs(head.x);
          que.push(head);
          fu++;
        }
      } else if (head.x >= 0) {
        if (fu % 2) {
          head.x += x;
          head.ab = abs(head.x);
          que.push(head);
          if (head.x < 0) fu++;
        } else {
          head.x -= x;
          head.ab = abs(head.x);
          que.push(head);
          if (head.x < 0) fu++;
        }
      }
    }
    while (!que.empty()) {
      ss aa = que.top();
      que.pop();
      ask[aa.id] = aa.x;
    }
    printf("%lld", ask[0]);
    for (int i = 1; i < n; i++) printf(" %lld", ask[i]);
    printf("\n");
  }
  return 0;
}
