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;
struct numb {
  long long z;
  int signal;
  int id;
  bool operator<(const numb &oth) const { return z < oth.z; }
};
int main() {
  int n, k, x;
  multiset<numb> s;
  cin >> n >> k >> x;
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    numb a;
    long long x;
    cin >> x;
    if (x < 0) {
      a.signal = -1;
      cnt++;
    } else
      a.signal = 1;
    a.z = abs(x);
    a.id = i;
    s.insert(a);
  }
  if (cnt % 2 == 0) {
    numb a = *s.begin();
    s.erase(s.begin());
    cnt = a.z / x + 1;
    a.z -= min(cnt, k) * x;
    k -= min(cnt, k);
    if (a.z < 0) {
      a.z = abs(a.z);
      a.signal *= -1;
    }
    s.insert(a);
  }
  for (int i = 0; i < k; ++i) {
    numb a = *s.begin();
    s.erase(s.begin());
    a.z += x;
    s.insert(a);
  }
  vector<long long> ans(n);
  for (auto it = s.begin(); it != s.end(); ++it) {
    ans[it->id] = it->z * it->signal;
  }
  for (int i = 0; i < n; ++i) {
    cout << ans[i] << " ";
  }
  return 0;
}
