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 = 200000 + 20;
long long aabs(long long x) { return x < 0 ? -x : x; }
struct node {
  long long num;
  int id;
  bool operator<(const node &temp) const { return aabs(num) > aabs(temp.num); }
};
long long n, k, x, a[N];
int main() {
  cin >> n >> k >> x;
  int sign = 1;
  priority_queue<node> Q;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", a + i);
    if (a[i] < 0) sign = -sign;
    Q.push((node){a[i], i});
  }
  while (k--) {
    node temp = Q.top();
    Q.pop();
    if (a[temp.id] < 0) {
      if (sign == -1)
        a[temp.id] -= x;
      else
        a[temp.id] += x;
      if (a[temp.id] >= 0) sign = -sign;
    } else {
      if (sign == -1)
        a[temp.id] += x;
      else
        a[temp.id] -= x;
      if (a[temp.id] < 0) sign = -sign;
    }
    Q.push((node){a[temp.id], temp.id});
  }
  for (int i = 1; i <= n; i++) printf("%lld%c", a[i], i == n ? '\n' : ' ');
  return 0;
}
