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;
void solve(vector<long long> &positive, int &k, long long x) {
  priority_queue<pair<long long, pair<long long, int>>,
                 vector<pair<long long, pair<long long, int>>>,
                 greater<pair<long long, pair<long long, int>>>>
      q;
  for (int i = 0; i < positive.size(); i++) {
    q.push(pair<long long, pair<long long, int>>(abs(positive[i]),
                                                 make_pair(positive[i], i)));
  }
  while (k-- > 0) {
    pair<long long, pair<long long, int>> u = q.top();
    q.pop();
    if (u.second.first < 0) {
      q.push(pair<long long, pair<long long, int>>(
          u.first + x, make_pair(u.second.first - x, u.second.second)));
    } else {
      q.push(pair<long long, pair<long long, int>>(
          u.first + x, make_pair(u.second.first + x, u.second.second)));
    }
  }
  while (!q.empty()) {
    pair<long long, pair<long long, int>> u = q.top();
    q.pop();
    positive[u.second.second] = u.second.first;
  }
}
int main() {
  int n, k;
  long long x;
  cin >> n >> k >> x;
  vector<long long> all(n);
  int negative = 0;
  for (int i = 0; i < n; i++) cin >> all[i], negative += all[i] < 0;
  int mn = -1, maxn = -INT_MAX, minp = INT_MAX, mx = -1;
  for (int i = 0; i < n; i++) {
    if (maxn < all[i] && all[i] < 0) {
      mx = i;
      maxn = all[i];
    }
    if (minp > all[i] && all[i] >= 0) {
      mn = i;
      minp = all[i];
    }
  }
  if (negative % 2 == 0) {
    if (abs(all[mx]) > abs(all[mn])) {
      while (all[mn] >= 0 && k) {
        k--;
        all[mn] -= x;
      }
    } else {
      while (all[mx] < 0 && k) {
        k--;
        all[mx] += x;
      }
    }
  }
  solve(all, k, x);
  for (int i = 0; i < n; i++) cout << all[i] << " ";
}
