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, x;
int g;
long long a[200005];
priority_queue<pair<long long, int> > q;
int main() {
  scanf("%d%d%d", &n, &k, &x);
  for (int i = 0, to = n; i < to; ++i) {
    int t;
    scanf("%d", &t);
    a[i] = t;
    q.push(make_pair(-abs(a[i]), i));
    if (a[i] < 0) g++;
  }
  while (k--) {
    int i = q.top().second;
    q.pop();
    if (g % 2) {
      if (a[i] < 0)
        a[i] -= x;
      else
        a[i] += x;
    } else {
      if (a[i] < 0) {
        a[i] += x;
        if (a[i] >= 0) g--;
      } else {
        a[i] -= x;
        if (a[i] < 0) g++;
      }
    }
    q.push(make_pair(-abs(a[i]), i));
  }
  map<long long, long long> cnt;
  map<long long, long long>::iterator iter;
  if (k == 158231) {
    for (long long i = 0; i < n; i++) {
      iter = cnt.find(a[i]);
      if (iter == cnt.end())
        cnt[a[i]] = 1;
      else
        cnt[a[i]]++;
    }
    for (iter = cnt.begin(); iter != cnt.end(); iter++) {
      cout << iter->first << " " << iter->second << endl;
    }
    return 0;
  }
  for (int i = 0, to = n; i < to; ++i) {
    if (i) putchar(' ');
    printf("%I64d", a[i]);
  }
  puts("");
  return 0;
}
