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;
int n, m, K, lst[200010];
long long ans[200010];
priority_queue<pair<long long, pair<long long, int> >,
               vector<pair<long long, pair<long long, int> > >,
               greater<pair<long long, pair<long long, int> > > >
    q;
int main() {
  int i;
  bool isNega = false;
  pair<long long, pair<long long, int> > tp;
  scanf("%d%d%d", &n, &K, &m);
  for (i = 1; i <= n; i++) {
    scanf("%d", &lst[i]);
    if (lst[i] < 0) isNega = !isNega;
    q.push(pair<long long, pair<long long, int> >(
        abs(lst[i]), pair<long long, int>(lst[i], i)));
  }
  while (K--) {
    tp = q.top();
    q.pop();
    if (!isNega) {
      if (tp.second.first >= 0) {
        tp.second.first -= m;
        if (tp.second.first < 0) isNega = !isNega;
      } else {
        tp.second.first += m;
        if (tp.second.first >= 0) isNega = !isNega;
      }
    } else {
      if (tp.second.first >= 0)
        tp.second.first += m;
      else
        tp.second.first -= m;
    }
    tp.first = abs(tp.second.first);
    q.push(tp);
  }
  while (!q.empty()) {
    ans[q.top().second.second] = q.top().second.first;
    q.pop();
  }
  for (i = 1; i <= n; i++) printf("%I64d ", ans[i]);
  return 0;
}
