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;
long long n, x, k;
struct node {
  long long v, id;
  node(long long a = 0, long long b = 0) : v(a), id(b) {}
  bool operator<(const node& y) const { return abs(v) > abs(y.v); }
} a[10000021];
priority_queue<node> q;
long long ans[1021141];
void add(node& u, long long op) {
  if (u.v >= 0)
    u.v += op * x;
  else
    u.v -= op * x;
}
int main() {
  scanf("%lld%lld%lld", &n, &k, &x);
  long long now = 1;
  for (long long z, i = 1; i <= n; i++) {
    scanf("%lld", &z);
    q.push(node(z, i));
    if (z < 0) now *= -1;
  }
  while (k--) {
    node u = q.top();
    q.pop();
    if (now == 1) {
      if (u.v < x) now *= -1;
      add(u, -1);
    } else
      add(u, 1);
    q.push(u);
  }
  while (!q.empty()) {
    node u = q.top();
    ans[u.id] = u.v;
    q.pop();
  }
  for (long long i = 1; i <= n; i++) printf("%lld ", ans[i]);
  return 0;
}
