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 N = 200100, MAX = 1000100;
struct Num {
  int i;
  long long val;
  bool operator<(const Num &a1) const { return abs(val) > abs(a1.val); }
} a[N];
int main() {
  int n, k, num1;
  long long x;
  while (~scanf("%d%d%lld", &n, &k, &x)) {
    priority_queue<Num> Q;
    num1 = 0;
    for (int i = 0; i < n; i++) {
      scanf("%lld", &a[i].val);
      a[i].i = i;
      Q.push(a[i]);
      if (a[i].val < 0) num1++;
    }
    Num ans;
    while (k--) {
      ans = Q.top();
      Q.pop();
      if (num1 % 2) {
        if (ans.val == 0) {
          ans.val += x;
        } else if (ans.val > 0) {
          ans.val += x;
        } else {
          ans.val -= x;
        }
      } else {
        if (ans.val == 0) {
          ans.val -= x;
          if (ans.val < 0) num1++;
        } else if (ans.val > 0) {
          ans.val -= x;
          if (ans.val < 0) num1++;
        } else {
          ans.val += x;
          if (ans.val >= 0) num1--;
        }
      }
      a[ans.i] = ans;
      Q.push(ans);
    }
    printf("%lld", a[0].val);
    for (int i = 1; i < n; i++) printf(" %lld", a[i].val);
    puts("");
  }
  return 0;
}
