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;
int main() {
  ios::sync_with_stdio(false);
  int n, k;
  long long x;
  cin >> n >> k >> x;
  vector<long long> a(n);
  vector<int> q(n);
  int nneg = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    q[i] = i;
    if (a[i] < 0) nneg++;
  }
  auto cmp = [&](int i, int j) { return llabs(a[i]) > llabs(a[j]); };
  make_heap(q.begin(), q.end(), cmp);
  while (k) {
    int i = q.front();
    pop_heap(q.begin(), q.end(), cmp);
    if (nneg % 2) {
      if (a[i] < 0) {
        a[i] -= x;
      } else {
        a[i] += x;
      }
    } else {
      if (a[i] < 0) {
        a[i] += x;
        if (a[i] >= 0) nneg--;
      } else {
        a[i] -= x;
        if (a[i] < 0) nneg++;
      }
    }
    k--;
    push_heap(q.begin(), q.end(), cmp);
  }
  cout << a[0];
  for (int i = 1; i < n; i++) cout << ' ' << a[i];
  cout << '\n';
}
