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 MAX_N = 2e5 + 9;
const int INF = 1e9 + 7;
long long N, M, T, S;
long long vec[MAX_N];
priority_queue<pair<long long, int>, vector<pair<long long, int> >,
               greater<pair<long long, int> > >
    que;
int main() {
  while (cin >> N >> M >> S) {
    while (!que.empty()) que.pop();
    int num = 0;
    for (int i = 0; i < N; i++) {
      scanf("%lld", &vec[i]);
      if (vec[i] < 0) num++;
      que.push(pair<long long, int>(abs(vec[i]), i));
    }
    if (num % 2 == 0) {
      pair<long long, int> p = que.top();
      que.pop();
      int x = p.second;
      if (vec[x] >= 0) {
        while (vec[x] >= 0 && M > 0) {
          vec[x] -= S;
          M -= 1;
        }
      } else {
        while (vec[x] < 0 && M > 0) {
          vec[x] += S;
          M -= 1;
        }
      }
      que.push(pair<long long, int>(abs(vec[x]), x));
    }
    for (int i = 0; i < M; i++) {
      pair<long long, int> p = que.top();
      que.pop();
      int x = p.second;
      if (vec[x] >= 0) {
        vec[x] += S;
      } else {
        vec[x] -= S;
      }
      que.push(pair<long long, int>(abs(vec[x]), x));
    }
    for (int i = 0; i < N; i++) {
      printf("%lld ", vec[i]);
    }
    cout << endl;
  }
  return 0;
}
