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(2e5 + 10);
int N, K, x, C;
long long A[maxN], ans[maxN];
priority_queue<pair<long long, int>, vector<pair<long long, int> >,
               greater<pair<long long, int> > >
    heap;
template <typename T>
T Abs(const T x) {
  return x > 0 ? x : -x;
}
int main() {
  cin >> N >> K >> x;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    ans[i] = A[i];
    C += A[i] < 0;
    heap.push(pair<long long, int>(Abs(A[i]), i));
  }
  while (!heap.empty() && K) {
    const pair<long long, int> top(heap.top());
    heap.pop();
    const long long val(top.first), i(top.second);
    if (C % 2) {
      if (ans[i] < 0)
        ans[i] -= x;
      else
        ans[i] += x;
    } else {
      if (ans[i] < 0) {
        ans[i] += x;
        C += ans[i] >= 0;
      } else {
        ans[i] -= x;
        C += ans[i] < 0;
      }
    }
    K--;
    heap.push(pair<long long, int>(Abs(ans[i]), i));
  }
  for (int i = 0; i < N; i++) cout << ans[i] << ' ';
}
