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;
int n, k, x, cnt;
long long a[200005];
struct Node {
  int id;
  long long val;
  Node() {}
  Node(int _id, long long _val) : id(_id), val(_val) {}
  bool operator<(const Node& rhs) const { return abs(val) > abs(rhs.val); }
};
int main() {
  scanf("%d%d%d", &n, &k, &x);
  priority_queue<Node> Q;
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &a[i]);
    if (a[i] < 0) cnt++;
    Q.push(Node(i, a[i]));
  }
  for (int i = 0; i < k; i++) {
    Node e = Q.top();
    long long t = e.val;
    Q.pop();
    if (cnt & 1) {
      if (e.val < 0)
        e.val -= x;
      else
        e.val += x;
    } else {
      if (e.val < 0)
        e.val += x;
      else
        e.val -= x;
    }
    if (e.val < 0 && t >= 0) cnt++;
    if (e.val >= 0 && t < 0) cnt--;
    a[e.id] = e.val;
    Q.push(e);
  }
  for (int i = 0; i < n; i++) printf("%I64d ", a[i]);
  return 0;
}
