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 a[300000];
set<pair<long long, long long> > s;
int main() {
  long long n, k, x, z = 0, neg = 0, ind = 0;
  cin >> n >> k >> x;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    s.insert(make_pair(abs(a[i]), i));
    if (a[i] == 0) {
      z++;
      ind = i;
    }
    if (a[i] < 0) neg++;
  }
  if (z > k) {
    a[0] -= k * x;
    for (long long i = 0; i < n; i++) cout << a[i] << " ";
    return 0;
  }
  if (z == 0 && neg % 2 == 0) {
    pair<long long, long long> h = *s.begin();
    if (a[h.second] > 0)
      while (a[h.second] >= 0 && k > 0) {
        a[h.second] -= x;
        k--;
      }
    else
      while (a[h.second] <= 0 && k > 0) {
        a[h.second] += x;
        k--;
      }
    s.erase(h);
    s.insert(make_pair(abs(a[h.second]), h.second));
  }
  for (long long i = 0; i < k; i++) {
    pair<long long, long long> tmp = *s.begin();
    if (a[tmp.second] <= 0)
      a[tmp.second] -= x;
    else
      a[tmp.second] += x;
    s.erase(tmp);
    s.insert(make_pair(abs(a[tmp.second]), tmp.second));
  }
  if ((z + neg) % 2 == 0 && neg != 0 && z != 0) {
    a[ind] *= (-1);
  }
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
  return 0;
}
