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 int MAXN = 200000;
long long n, k, x, a[MAXN + 10];
priority_queue<pair<long long, pair<int, long long> >,
               vector<pair<long long, pair<int, long long> > >,
               greater<pair<long long, pair<int, long long> > > >
    que;
long long _abs(const long long u) { return u > 0 ? u : -u; }
int main() {
  cin >> n >> k >> x;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int mode = 0;
  for (int i = 1; i <= n; i++) {
    que.push(make_pair(_abs(a[i]), make_pair(i, a[i])));
    if (a[i] < 0) mode ^= 1;
  }
  long long pre;
  while (k--) {
    pair<long long, pair<int, long long> > u = que.top();
    que.pop();
    pre = u.second.second;
    if (mode) {
      if (u.second.second < 0)
        u.second.second -= x;
      else
        u.second.second += x;
    } else {
      if (u.second.second < 0)
        u.second.second += x;
      else
        u.second.second -= x;
    }
    if ((u.second.second < 0) ^ (pre < 0)) mode ^= 1;
    u.first = _abs(u.second.second);
    que.push(u);
  }
  while (!que.empty()) {
    a[que.top().second.first] = que.top().second.second;
    que.pop();
  }
  for (int i = 1; i <= n; i++) {
    cout << a[i];
    if (i != n) printf(" ");
  }
  return 0;
}
