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;
long long pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
int n, k;
long long x, arr[200005];
void func() {
  priority_queue<pair<long long, long long> > pq;
  for (int i = 1; i <= n; i++) {
    if (arr[i] >= 0)
      pq.push(pair<long long, long long>(-arr[i], i));
    else
      pq.push(pair<long long, long long>(arr[i], i));
  }
  while (k--) {
    int p = pq.top().second;
    pq.pop();
    if (arr[p] >= 0)
      arr[p] += x;
    else
      arr[p] -= x;
    if (arr[p] >= 0)
      pq.push(pair<long long, long long>(-arr[p], p));
    else
      pq.push(pair<long long, long long>(arr[p], p));
  }
  for (int i = 1; i <= n; i++) printf("%lld ", arr[i]);
}
int main() {
  scanf("%d%d%lld", &n, &k, &x);
  int p = 1, negs = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &arr[i]);
    if (abs(arr[i]) < abs(arr[p])) p = i;
    if (arr[i] < 0) negs++;
  }
  if (negs % 2 == 0) {
    if (arr[p] >= 0) {
      while (k > 0 && arr[p] >= 0) {
        k--;
        arr[p] -= x;
      }
    } else {
      while (k > 0 && arr[p] <= 0) {
        k--;
        arr[p] += x;
      }
    }
  }
  func();
  return 0;
}
