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;
long long int n, p, m, k, x;
long long int a[200005 + 2] = {0};
void solve() {
  long long int c = 0;
  a[0] = 10000000000;
  long long int sexyass = 0;
  for (long long int i = 1; i <= n; i++) {
    if (a[i] < 0) c++;
    if (abs(a[i]) < abs(a[sexyass])) sexyass = i;
  }
  if (c % 2 == 0) {
    long long int kl = abs(a[sexyass]) / x + 1;
    kl = min(k, kl);
    k -= kl;
    if (a[sexyass] < 0)
      a[sexyass] += kl * x;
    else
      a[sexyass] -= kl * x;
  }
  set<pair<long long int, long long int> > s;
  for (long long int i = 1; i <= n; i++) {
    s.insert({abs(a[i]), i});
  }
  while (k > 0) {
    k -= 1;
    long long int p = s.begin()->second;
    if (a[p] >= 0)
      a[p] += x;
    else
      a[p] -= x;
    s.erase(s.begin());
    s.insert({abs(a[p]), p});
  }
  for (long long int i = 1; i <= n; i++) cout << a[i] << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k >> x;
  for (long long int i = 1; i <= n; i++) cin >> a[i];
  solve();
}
