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;
long long first;
int f[200200];
long long ans[200200];
priority_queue<pair<long long, int>, vector<pair<long long, int> >,
               greater<pair<long long, int> > >
    q;
int main() {
  int num = 0;
  long long t;
  pair<long long, int> u;
  scanf("%d%d%lld", &n, &k, &first);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &t);
    if (t >= 0)
      f[i] = 1;
    else
      f[i] = -1, num++;
    q.push(make_pair(abs(t), i));
  }
  for (int j = 0; j < k; j++) {
    u = q.top();
    q.pop();
    if (num & 1) {
      u.first += first;
    } else {
      u.first -= first;
      if (f[u.second] == 1)
        if (u.first < 0)
          f[u.second] = -1, u.first = -u.first, num++;
        else
          ;
      else if (u.first <= 0)
        f[u.second] = 1, u.first = -u.first, num--;
    }
    q.push(u);
  }
  for (int i = 0; i < n; i++) {
    u = q.top();
    q.pop();
    ans[u.second] = u.first * f[u.second];
  }
  for (int i = 0; i < n; i++) printf("%lld%c", ans[i], "\n "[i < n - 1]);
}
