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;
priority_queue<pair<pair<long long int, int>, int>,
               vector<pair<pair<long long int, int>, int> >,
               greater<pair<pair<long long int, int>, int> > >
    q;
pair<pair<long long int, int>, int> temp;
long long int ans[200010];
int main() {
  ios_base::sync_with_stdio(false);
  int n, k, neg = 0;
  long long int x, a;
  cin >> n >> k >> x;
  for (int i = 0; i < n; i++) {
    cin >> a;
    if (a < 0) {
      neg++;
      q.push(make_pair(make_pair(abs(a), 0), i));
    } else
      q.push(make_pair(make_pair(abs(a), 1), i));
  }
  if (neg % 2 == 0) {
    temp = q.top();
    q.pop();
    while (temp.first.first >= 0 && k) {
      k--;
      temp.first.first -= x;
    }
    if (temp.first.first >= 0)
      q.push(temp);
    else
      q.push(make_pair(make_pair(abs(temp.first.first), 1 - temp.first.second),
                       temp.second));
  }
  while (k) {
    k--;
    temp = q.top();
    q.pop();
    q.push(make_pair(make_pair(temp.first.first + x, temp.first.second),
                     temp.second));
  }
  while (!q.empty()) {
    temp = q.top();
    q.pop();
    if (temp.first.second == 0)
      ans[temp.second] = (-1) * temp.first.first;
    else
      ans[temp.second] = temp.first.first;
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << "\n";
  return 0;
}
