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 n, k, x, dem, ans, f[200005], has_0;
priority_queue<pair<long long, long>, vector<pair<long long, long> >,
               greater<pair<long long, long> > >
    q;
vector<pair<long long, long> > a;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k >> x;
  for (long long i = 1; i <= n; i++) {
    cin >> f[i];
    if (f[i] < 0) dem++;
    if (f[i] == 0) has_0++;
    q.push(pair<long long, long>(abs(f[i]), i));
  }
  if (has_0 > k) {
    f[1] += k * x;
    for (long long i = 1; i <= n; i++) cout << f[i] << ' ';
    return 0;
  }
  if (dem % 2 == 0) {
    pair<long long, long> A = q.top();
    q.pop();
    long long id = A.second;
    if (f[id] >= 0)
      while (f[id] >= 0 && k > 0) f[id] -= x, k--;
    else
      while (f[id] < 0 && k > 0) f[id] += x, k--;
    q.push(pair<long long, long>(abs(f[id]), id));
  }
  while (k > 0) {
    k--;
    pair<long long, long> A = q.top();
    q.pop();
    long long v = A.first, id = A.second;
    v += x;
    if (f[id] < 0)
      f[id] = -v;
    else if (f[id] >= 0)
      f[id] = v;
    q.push(pair<long long, long>(v, id));
  }
  for (long long i = 1; i <= n; i++) cout << f[i] << ' ';
  return 0;
}
