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;
struct llabss {
  int idx;
  long long x;
};
bool operator<(const llabss& a, const llabss& b) { return abs(a.x) > abs(b.x); }
priority_queue<llabss> que;
long long n, k, x;
long long res[222222];
bool nega = false;
int main() {
  cin.sync_with_stdio(false);
  cin >> n >> k >> x;
  for (int i = (int)(0); i < (int)(n); i++) {
    long long a;
    cin >> a;
    que.push((llabss){i, a});
    if (a < 0) {
      if (nega)
        nega = false;
      else
        nega = true;
    }
  }
  for (int hoge = (int)(0); hoge < (int)(k); hoge++) {
    llabss ns = que.top();
    que.pop();
    int idx = ns.idx;
    long long xx = ns.x;
    if (nega) {
      if (xx < 0)
        xx -= x;
      else
        xx += x;
    } else {
      if (xx >= 0) {
        xx -= x;
        if (xx < 0) {
          if (nega)
            nega = false;
          else
            nega = true;
        }
      } else {
        xx += x;
        if (xx >= 0) {
          if (nega)
            nega = false;
          else
            nega = true;
        }
      }
    }
    que.push((llabss){idx, xx});
  }
  for (int i = (int)(0); i < (int)(n); i++) {
    llabss ns = que.top();
    que.pop();
    res[ns.idx] = ns.x;
  }
  cout << res[0];
  for (int i = (int)(1); i < (int)(n); i++) cout << " " << res[i];
  cout << endl;
  return 0;
}
