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, pos, neg;
long long arr[200005];
multiset<pair<long long, int>> ms;
multiset<pair<long long, int>>::iterator it1, it2;
bool comp(pair<long long, int> pp, pair<long long, int> b) {
  return pp.first < b.first;
}
int main() {
  cin >> n >> k >> x;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    ms.insert({abs(arr[i]), i});
    if (arr[i] < 0) neg++;
  }
  pair<long long, int> pp;
  if (neg % 2 == 0) {
    it1 = ms.begin();
    pp = *it1;
    while (k > 0) {
      if (arr[pp.second] < 0) {
        arr[pp.second] += x;
        k--;
        if (arr[pp.second] >= 0) break;
      } else {
        arr[pp.second] -= x;
        k--;
        if (arr[pp.second] < 0) break;
      }
    }
    ms.erase(*ms.begin());
    ms.insert({abs(arr[pp.second]), pp.second});
  }
  while (k > 0) {
    it1 = ms.begin();
    pp = *it1;
    if (arr[pp.second] < 0) {
      arr[pp.second] -= x;
    } else {
      arr[pp.second] += x;
    }
    k--;
    ms.erase(*ms.begin());
    ms.insert({abs(arr[pp.second]), pp.second});
  }
  for (int i = 0; i < n; i++) cout << arr[i] << " ";
  return 0;
}
