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(void) {
  srand(time(0));
  cout << fixed << setprecision(7);
  cerr << fixed << setprecision(7);
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  int n, k, m;
  cin >> n >> k >> m;
  multiset<pair<pair<long long, int>, int> > Q;
  int ng = 0;
  for (int i = 1; i <= n; ++i) {
    int v;
    cin >> v;
    ng += v < 0;
    Q.insert({{abs(v), (bool)(v < 0)}, i});
  }
  static long long ans[1 << 20];
  while (k--) {
    auto node = *Q.begin();
    Q.erase(Q.begin());
    long long v = node.first.first * (node.first.second ? -1 : 1);
    ng -= node.first.second;
    if (ng & 1)
      v += m;
    else
      v -= m;
    ng += v < 0;
    node.first = {abs(v), (bool)(v < 0)};
    Q.insert(node);
  }
  for (auto it : Q)
    ans[it.second] = it.first.first * (it.first.second ? -1 : 1);
  for (int i = 1; i <= n; ++i) cout << ans[i] << " \n"[i == n];
  cerr << "Time elapsed :" << clock() * 1000.0 / CLOCKS_PER_SEC << " ms"
       << '\n';
  return 0;
}
