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>
long long mod = 1000000007;
using namespace std;
long long a[200200], s[200200];
map<pair<long long, long long>, long long> mp;
map<pair<long long, long long>, long long>::iterator it;
long long n, k, x, pos = 0, neg = 0, zero = 0, old, mn = 1e10, i, mx = -1e10;
int main() {
  cin >> n >> k >> x;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] > 0)
      pos++;
    else if (a[i] < 0)
      neg++;
    else
      zero++;
    if (a[i] >= 0)
      s[i] = 1;
    else
      s[i] = -1;
    mp[make_pair(abs(a[i]), i)]++;
  }
  if (neg % 2 == 0) {
    for (i = 0; i < n; i++)
      if (mn > a[i] && a[i] >= 0) mn = a[i];
    for (i = 0; i < n; i++)
      if (mx < a[i] && a[i] < 0) mx = a[i];
    if (abs(mx) >= abs(mn)) {
      old = mn;
      while (mn >= 0 && k != 0) {
        mn = mn - x;
        k--;
      }
      for (i = 0; i < n; i++)
        if (old == a[i]) {
          mp.erase(make_pair(abs(a[i]), i));
          a[i] = mn;
          mp[make_pair(abs(a[i]), i)]++;
          if (a[i] >= 0)
            s[i] = 1;
          else
            s[i] = -1;
          break;
        }
    } else {
      old = mx;
      while (mx <= 0 && k != 0) {
        mx = mx + x;
        k--;
      }
      for (i = 0; i < n; i++)
        if (old == a[i]) {
          mp.erase(make_pair(abs(a[i]), i));
          a[i] = mx;
          mp[make_pair(abs(a[i]), i)]++;
          if (a[i] > 0)
            s[i] = 1;
          else if (a[i] < 0)
            s[i] = -1;
          else
            s[i] = 0;
          break;
        }
    }
  }
  while (k != 0) {
    it = mp.begin();
    mp[make_pair((*it).first.first + x, (*it).first.second)]++;
    mp.erase(it);
    k--;
  }
  for (it = mp.begin(); it != mp.end(); it++)
    a[(*it).first.second] = (*it).first.first * s[(*it).first.second];
  for (i = 0; i < n; i++) cout << a[i] << " ";
  return 0;
}
