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 long long INF = 1e12;
int n, k;
long long x;
long long a[200010];
void absmax(void) {
  priority_queue<pair<long long, int>, vector<pair<long long, int> >,
                 greater<pair<long long, int> > >
      q;
  for (int i = 0; i < (n); i++) {
    q.push(make_pair(abs(a[i]), i));
  }
  for (int i = 0; i < (k); i++) {
    auto tmp = q.top();
    q.pop();
    int idx = tmp.second;
    if (a[idx] >= 0) {
      a[idx] += x;
      q.push(make_pair(abs(a[idx]), idx));
    } else {
      a[idx] -= x;
      q.push(make_pair(abs(a[idx]), idx));
    }
  }
  return;
}
int main(void) {
  cin >> n >> k >> x;
  long long sign = 0;
  long long mi = INF;
  int miidx;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    if (a[i] < 0) sign++;
    if (abs(a[i]) < abs(mi)) {
      mi = a[i];
      miidx = i;
    }
  }
  sign %= 2;
  if (sign == 1) {
    absmax();
  } else {
    if (abs(mi) < k * x) {
      if (mi >= 0) {
        while (a[miidx] >= 0) {
          a[miidx] -= x;
          k--;
        }
        absmax();
      } else {
        while (a[miidx] <= 0) {
          a[miidx] += x;
          k--;
        }
        absmax();
      }
    } else {
      if (mi >= 0) {
        a[miidx] -= k * x;
      } else {
        a[miidx] += k * x;
      }
    }
  }
  for (int i = 0; i < (n); i++) {
    printf("%lld ", a[i]);
  }
  printf("\n");
}
