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 sign(long long x) {
  if (x > 0) {
    return 1;
  } else if (x < 0) {
    return -1;
  }
  return 0;
}
void solve(vector<long long>& xs, long long k, long long x) {
  priority_queue<pair<long long, long long>, vector<pair<long long, long long>>,
                 greater<pair<long long, long long>>>
      que;
  long long sgn = 1;
  for (long long i = 0; i < xs.size(); ++i) {
    que.push({abs(xs[i]), i});
    if (xs[i] < 0) {
      sgn *= -1;
    }
  }
  if (sgn == 1) {
    auto t = que.top();
    que.pop();
    long long i = t.second;
    if (xs[i] >= 0) {
      while (k > 0 && xs[i] >= 0) {
        xs[i] -= x;
        --k;
      }
    } else {
      while (k > 0 && xs[i] <= 0) {
        xs[i] += x;
        --k;
      }
    }
    que.push({abs(xs[i]), i});
  }
  while (k > 0) {
    auto t = que.top();
    que.pop();
    auto i = t.second;
    xs[i] += (xs[i] >= 0 ? x : -x);
    que.push({abs(xs[i]), i});
    --k;
  }
}
int main() {
  while (true) {
    long long n, k, x;
    cin >> n >> k >> x;
    if (!cin) {
      break;
    }
    vector<long long> xs(n, 0);
    for (long long i = 0; i < n; ++i) {
      cin >> xs[i];
    }
    solve(xs, k, x);
    for (long long i = 0; i < xs.size(); ++i) {
      cout << xs[i] << " ";
    }
    cout << endl;
  }
  return 0;
}
