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 node {
  long long a;
  int id;
} s[2000000];
bool operator<(node x, node y) { return fabs(x.a) > fabs(y.a); }
priority_queue<node> q;
long long b[200333];
int main() {
  int k, x;
  int n;
  scanf("%d", &n);
  scanf("%d%d", &k, &x);
  int count = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &s[i].a);
    s[i].id = i;
    if (s[i].a < 0) count++;
  }
  for (int i = 1; i <= n; i++) {
    q.push(s[i]);
  }
  while (k != 0) {
    k--;
    node temp = q.top();
    q.pop();
    if (count & 1) {
      if (temp.a >= 0) {
        temp.a += x;
        q.push(temp);
      } else {
        temp.a -= x;
        q.push(temp);
      }
    } else {
      if (temp.a >= 0) {
        temp.a -= x;
        if (temp.a < 0) count++;
        q.push(temp);
      } else {
        temp.a += x;
        if (temp.a >= 0) count++;
        q.push(temp);
      }
    }
  }
  while (!q.empty()) {
    node temp = q.top();
    q.pop();
    b[temp.id] = temp.a;
  }
  for (int i = 1; i <= n; i++) {
    if (i == 1)
      printf("%I64d", b[i]);
    else
      printf(" %I64d", b[i]);
  }
  printf("\n");
}
