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 a[200200];
struct comp {
  bool operator()(int x, int y) {
    if (abs(a[x]) == abs(a[y])) return (x < y);
    return (abs(a[x]) < abs(a[y]));
  }
};
set<int, comp> st;
int n = 0, k, x, kk;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k >> x;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] < 0) kk++;
    st.insert(i);
  }
  if (kk % 2 == 0) {
    int i = *st.begin();
    int m = (a[i] ? a[i] / abs(a[i]) : 1);
    a[i] = abs(a[i]);
    int step = a[i] / x + 1;
    if (step <= k) {
      a[i] = x - a[i] % x;
      a[i] *= -m;
      k -= step;
    } else
      a[i] -= x * k, a[i] *= m, k = 0;
    st.erase(st.begin());
    st.insert(i);
  }
  while (k--) {
    int i = *st.begin();
    int m = (a[i] ? a[i] / abs(a[i]) : 1);
    a[i] = abs(a[i]);
    a[i] += x;
    a[i] *= m;
    st.erase(st.begin());
    st.insert(i);
  }
  for (int i = 0; i < n; i++) cout << a[i] << ' ';
}
