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 nm = 200010;
int n;
long long k, x;
long long a[nm];
int main() {
  scanf("%d%I64d%I64d", &n, &k, &x);
  int d0 = 0, dam = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", &a[i]);
    if (a[i] == 0)
      d0++;
    else if (a[i] < 0)
      dam++;
  }
  if (d0) {
    if (d0 > k) {
      for (int i = 1; i <= n; ++i) cout << a[i] << " ";
      return 0;
    }
    for (int i = 1; i <= n; ++i) {
      if (a[i] == 0) {
        if (d0 > 1) {
          a[i] += x;
          k--;
          d0--;
        } else {
          if (dam % 2) {
            a[i] += x;
            k--;
            d0--;
          } else {
            a[i] -= x;
            k--;
            d0--;
            dam++;
          }
        }
      }
    }
  }
  if (dam % 2 == 0) {
    int ri = 1;
    for (int i = 2; i <= n; ++i)
      if (abs(a[i]) < abs(a[ri])) ri = i;
    long long tmp = a[ri];
    for (int i = 1; k > 0 && tmp * a[ri] >= 0; ++i) {
      if (tmp > 0)
        a[ri] -= x;
      else
        a[ri] += x;
      k--;
    }
  }
  set<pair<long long, int> > xxx;
  for (int i = 1; i <= n; ++i) {
    xxx.insert(make_pair(abs(a[i]), i));
  }
  for (int i = 1; i <= k; ++i) {
    int j = xxx.begin()->second;
    if (a[j] < 0)
      a[j] -= x;
    else
      a[j] += x;
    xxx.erase(xxx.begin());
    xxx.insert(make_pair(abs(a[j]), j));
  }
  for (int i = 1; i <= n; ++i) cout << a[i] << " ";
}
