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;
const int N = 2e5 + 10;
long long int a[N];
long long int n, k, x;
set<pair<long long int, int> > s;
int main() {
  cin >> n >> k >> x;
  for (int i = 0; i < n; ++i) cin >> a[i];
  int num = 0;
  for (int i = 0; i < n; ++i)
    if (a[i] < 0) num++;
  for (int i = 0; i < n; ++i) s.insert(make_pair(abs(a[i]), i));
  pair<long long int, int> now;
  for (int i = 0; i < k; ++i) {
    now = *(s.begin());
    s.erase(s.begin());
    num -= (a[now.second] < 0);
    if (num % 2) a[now.second] += x;
    if (!(num % 2)) a[now.second] -= x;
    num += (a[now.second] < 0);
    s.insert(make_pair(abs(a[now.second]), now.second));
  }
  for (int i = 0; i < n; ++i) cout << a[i] << ' ';
}
