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 n, k, x;
vector<long long> a;
void ans() {
  for (int i = 0; i < n; i++) {
    cout << a[i] << ' ';
  }
  exit(0);
}
int main(int argc, char* argv[]) {
  srand((unsigned int)time(NULL));
  cin >> n >> k >> x;
  a.resize(n);
  int ch = 1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] < 0) {
      ch *= -1;
    }
  }
  if (ch == 1) {
    set<pair<long long, int> > s;
    for (int i = 0; i < n; i++) {
      if (a[i] >= 0) {
        s.insert({abs(-1 - a[i]), i});
      } else {
        s.insert({abs(1 - a[i]), i});
      }
    }
    auto q = *s.begin();
    int id = q.second;
    if (a[id] >= 0) {
      while (a[id] >= 0) {
        a[id] -= x;
        k--;
        if (k == 0) {
          ans();
        }
      }
    } else {
      while (a[id] <= 0) {
        a[id] += x;
        k--;
        if (k == 0) {
          ans();
        }
      }
    }
  }
  set<pair<long long, int> > s;
  for (int i = 0; i < n; i++) {
    s.insert({abs(a[i]), i});
  }
  while (k > 0) {
    auto q = *s.begin();
    s.erase(s.begin());
    int id = q.second;
    if (a[id] >= 0) {
      a[id] += x;
    } else {
      a[id] -= x;
    }
    k--;
    s.insert({abs(a[id]), id});
  }
  ans();
}
