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 nod {
  long long num;
  int id;
  bool operator<(const nod& a) const {
    return (num < 0 ? -num : num) > (a.num < 0 ? -a.num : a.num);
  }
};
long long a[200100];
int n, k;
long long x;
int op = 1;
priority_queue<nod> q;
inline void get_ready() {
  cin >> n >> k >> x;
  for (int i = (1); i <= (n); i++) {
    scanf("%I64d", &a[i]);
    if (a[i] < 0) op = -op;
    q.push((nod){a[i], i});
  }
}
inline void solve() {
  while (k--) {
    nod t = q.top();
    q.pop();
    if (a[t.id] < 0) {
      if (op == 1)
        a[t.id] += x;
      else
        a[t.id] -= x;
      if (a[t.id] >= 0) op = -op;
    } else {
      if (op == 1)
        a[t.id] -= x;
      else
        a[t.id] += x;
      if (a[t.id] < 0) op = -op;
    }
    q.push((nod){a[t.id], t.id});
  }
  for (int i = (1); i <= (n); i++) printf("%I64d ", a[i]);
}
int main() {
  get_ready();
  solve();
  return 0;
}
