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 maxn = 2e5 + 5;
struct Node {
  long long val;
  int id;
  bool operator<(const Node& b) const { return abs(val) > abs(b.val); }
} a[maxn];
priority_queue<Node> q;
int main() {
  cin.sync_with_stdio(false);
  int n, k, x, cnt = 0, minn = 0;
  cin >> n >> k >> x;
  for (int i = 0; i < n; i++) {
    cin >> a[i].val;
    a[i].id = i;
    if (a[i].val < 0) cnt++;
    q.push(a[i]);
  }
  while (k--) {
    Node now = q.top();
    q.pop();
    if (now.val >= 0) {
      if (cnt & 1)
        a[now.id].val += x;
      else
        a[now.id].val -= x;
      if (a[now.id].val < 0) cnt++;
      q.push(a[now.id]);
    } else {
      if (cnt & 1)
        a[now.id].val -= x;
      else
        a[now.id].val += x;
      if (a[now.id].val >= 0) cnt--;
      q.push(a[now.id]);
    }
  }
  for (int i = 0; i < n; i++)
    printf("%I64d%c", a[i].val, i == n - 1 ? '\n' : ' ');
  return 0;
}
