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 mod = 1e9 + 7;
const int N = 1e6 + 100;
const long long inf = 1e18 + 100;
long double EPS = 1e-18;
const int INF = 2000000000;
int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};
int dxx[] = {0, 0, -1, -1, -1, 1, 1, 1};
int dyy[] = {1, -1, 0, 1, -1, 0, 1, -1};
bool valid(int x, int y, int nn, int mm) {
  return (x >= 0 && y >= 0 && x < nn && y < mm);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int n, k, x;
  cin >> n >> k >> x;
  long long arr[n], sgn = 1;
  set<pair<long long, int> > second;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] < 0) sgn *= -1;
    second.insert({abs(arr[i]), i});
  }
  while (k--) {
    int idx = second.begin()->second;
    second.erase(second.begin());
    if (arr[idx] < 0) sgn *= -1;
    arr[idx] -= sgn * x;
    if (arr[idx] < 0) sgn *= -1;
    second.insert({abs(arr[idx]), idx});
  }
  for (int i = 0; i < n; i++) cout << arr[i] << ' ';
  return 0;
}
