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 N = 200200;
int n, k;
long long a[N];
int zeroes = 0;
int sgn = 1;
long long m;
set<pair<long long, int> > setik;
int main() {
  scanf("%d%d%lld", &n, &k, &m);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    if (a[i] == 0) zeroes++;
    if (a[i] < 0) sgn *= -1;
  }
  if (zeroes > k) {
    for (int i = 0; i < n; i++) printf("%lld ", a[i]);
    printf("\n");
    return 0;
  }
  if (zeroes > 0) {
    k -= zeroes;
    for (int i = 0; i < n; i++) {
      if (a[i] != 0) continue;
      zeroes--;
      if (zeroes > 0) {
        a[i] = m;
        continue;
      }
      a[i] = -m * sgn;
      sgn = -1;
    }
  }
  if (sgn == 1) {
    int pos = -1;
    for (int i = 0; i < n; i++)
      if (pos == -1 || abs(a[i]) < abs(a[pos])) pos = i;
    if (a[pos] > 0) {
      while (k > 0 && a[pos] >= 0) {
        k--;
        a[pos] -= m;
      }
    } else {
      while (k > 0 && a[pos] <= 0) {
        k--;
        a[pos] += m;
      }
    }
  }
  for (int i = 0; i < n; i++) setik.insert(make_pair(abs(a[i]), i));
  while (k--) {
    int p = setik.begin()->second;
    setik.erase(setik.begin());
    if (a[p] > 0)
      a[p] += m;
    else
      a[p] -= m;
    setik.insert(make_pair(abs(a[p]), p));
  }
  for (int i = 0; i < n; i++) printf("%lld ", a[i]);
  printf("\n");
  return 0;
}
