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 n, k, x;
long long a[200005];
long long b[200005];
bool neg[200005];
void solve() {
  cin >> n >> k >> x;
  long long numNeg = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] < 0) {
      neg[i] = true;
      numNeg++;
    }
  }
  long long mi = 1000000000000000LL, mii = -1;
  for (long long i = 0; i < n; i++) {
    if (abs(a[i]) < mi) {
      mi = abs(a[i]);
      mii = i;
    }
  }
  if (numNeg % 2 == 0) {
    if (!neg[mii]) {
      while (k > 0) {
        if (a[mii] < 0) {
          break;
        }
        a[mii] -= x;
        k--;
      }
      if (a[mii] < 0) neg[mii] = true;
    } else {
      while (k > 0) {
        if (a[mii] > 0) {
          break;
        }
        a[mii] += x;
        k--;
      }
      if (a[mii] > 0) neg[mii] = false;
    }
  }
  for (long long i = 0; i < n; i++) b[i] = abs(a[i]);
  set<pair<long long, long long> > s;
  for (long long i = 0; i < n; i++) s.insert({b[i], i});
  while (k > 0) {
    pair<long long, long long> p = *s.begin();
    s.erase(s.begin());
    b[p.second] += x;
    s.insert({b[p.second], p.second});
    k--;
  }
  for (long long i = 0; i < n; i++) {
    if (neg[i])
      cout << -1 * b[i] << " ";
    else
      cout << b[i] << " ";
  }
  cout << endl;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
}
