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 INF = 1e9;
struct Node {
  long long x, bj, id;
  bool operator<(const Node &a) const { return x > a.x; }
} dq;
int n, k, x;
long long a[200005];
int s, id;
int main() {
  scanf("%d%d%d", &n, &k, &x);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &a[i]);
  s = 0;
  for (int i = 1; i <= n; ++i)
    if (a[i] < 0) s++;
  if (s % 2 == 0) {
    id = 1;
    for (int i = 1; i <= n; ++i)
      if (abs(a[i]) < abs(a[id])) id = i;
    if (a[id] < 0) {
      while (k > 0 && a[id] <= 0) {
        k--;
        a[id] += x;
      }
    } else {
      while (k > 0 && a[id] >= 0) {
        k--;
        a[id] -= x;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 0 && k > 0) {
      a[i] += x;
      k--;
    }
  }
  priority_queue<Node> q;
  for (int i = 1; i <= n; ++i) {
    if (a[i] >= 0) {
      dq.x = a[i];
      dq.bj = 1;
      dq.id = i;
    } else {
      dq.x = -a[i];
      dq.bj = -1;
      dq.id = i;
    }
    q.push(dq);
  }
  for (int i = 1; i <= k; ++i) {
    dq = q.top();
    q.pop();
    if (dq.bj == 1) {
      a[dq.id] += x;
      dq.x += x;
      q.push(dq);
    } else {
      a[dq.id] -= x;
      dq.x += x;
      q.push(dq);
    }
  }
  for (int i = 1; i <= n; ++i) printf("%I64d ", a[i]);
  return 0;
}
