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;
template <typename T>
bool mycomp(T x, T y) {
  return (x == y);
}
bool paircomp(const pair<long long int, long long int> &x,
              const pair<long long int, long long int> &y) {
  return x.second < y.second;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, k, x;
  cin >> n >> k >> x;
  long long int arr[n];
  for (long long int i = 0; i < n; i++) cin >> arr[i];
  ;
  set<pair<long long int, long long int> > sp;
  long long int cntneg = 0;
  for (long long int i = 0; i < n; i++) {
    sp.insert(make_pair(abs(arr[i]), i));
    if (arr[i] < 0) cntneg++;
  }
  long long int roxa = k;
  long long int flagr;
  while (k > 0) {
    pair<long long int, long long int> nms = *sp.begin();
    sp.erase(nms);
    long long int kk = arr[nms.second];
    if (arr[nms.second] < 0) {
      if (cntneg % 2 == 0)
        arr[nms.second] += x;
      else
        arr[nms.second] -= x;
    } else if (arr[nms.second] > 0) {
      if (cntneg % 2 == 0)
        arr[nms.second] -= x;
      else
        arr[nms.second] += x;
    } else {
      if (cntneg % 2 != 0)
        arr[nms.second] += x;
      else {
        if (roxa == k)
          arr[nms.second] -= x;
        else {
          if (flagr > 0)
            arr[nms.second] -= x;
          else {
            arr[nms.second] += x;
            cntneg--;
          }
        }
      }
    }
    flagr = kk;
    if (arr[nms.second] > 0 && kk < 0)
      cntneg--;
    else if (arr[nms.second] < 0 && kk >= 0)
      cntneg++;
    sp.insert(make_pair(abs(arr[nms.second]), nms.second));
    k--;
  }
  for (long long int i = 0; i < n; i++) cout << arr[i] << " ";
}
