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 + 10;
set<pair<long long, int> > sp;
int sign = 1;
int main() {
  long long n, k, x;
  scanf("%lld %lld %lld", &n, &k, &x);
  long long a[n];
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    if (a[i] < 0) sign *= -1;
    sp.insert({abs(a[i]), i});
  }
  while (k--) {
    pair<long long, int> num = *sp.begin();
    sp.erase(num);
    int index = num.second;
    if (a[index] < 0) sign *= -1;
    if (sign == 1)
      a[index] -= x;
    else
      a[index] += x;
    if (a[index] < 0) sign *= -1;
    sp.insert({abs(a[index]), index});
  }
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << endl;
  return 0;
}
