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 mxN = 2e5;
bool sgn[mxN + 5];
int N;
long long K, X;
int oddCt;
long long ans[mxN + 5];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N >> K >> X;
  priority_queue<pair<long long, long long>, vector<pair<long long, long long>>,
                 greater<pair<long long, long long>>>
      pq;
  for (int i = 1, x; i <= N; i++) {
    cin >> x;
    pq.push({abs(x), i});
    if (x < 0) oddCt++;
    sgn[i] = (x < 0);
  }
  if (oddCt % 2 == 0) {
    long long x = pq.top().first;
    long long id = pq.top().second;
    pq.pop();
    long long need = (x + X - 1) / X;
    long long mn = min(need, K);
    x -= mn * X;
    K -= mn;
    if (x >= 0 && K > 0) {
      K--;
      x -= X;
    }
    if (x < 0) {
      sgn[id] = !sgn[id];
      oddCt++;
    }
    pq.push({abs(x), id});
  }
  if (oddCt % 2 == 1) {
    while (K > 0) {
      pair<long long, long long> p = pq.top();
      pq.pop();
      pq.push({p.first + X, p.second});
      K--;
    }
  }
  while (!pq.empty()) {
    pair<long long, long long> p = pq.top();
    pq.pop();
    ans[p.second] = p.first;
    if (sgn[p.second]) ans[p.second] *= -1;
  }
  for (int i = 1; i <= N; i++) cout << ans[i] << (i == N ? "\n" : " ");
  return 0;
}
