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 TAM = 2e5 + 100;
struct compare {
  bool operator()(pair<long long, int>& x, pair<long long, int>& y) {
    if (abs(x.first) != abs(y.first)) return abs(x.first) > abs(y.first);
    return x.second < y.second;
  }
};
long long n, k, X;
long long a[TAM];
int main() {
  priority_queue<pair<long long, int>, vector<pair<long long, int> >, compare>
      pq;
  cin >> n >> k >> X;
  int x;
  int sgn = 1;
  int zeros = 0;
  for (int i = 0; i < n; i++) {
    cin >> x;
    pq.push({x, i});
    if (x != 0)
      sgn *= x / abs(x);
    else
      zeros++;
  }
  pair<long long, int> y;
  if (zeros) {
    if (sgn > 0) {
      y = pq.top();
      pq.pop();
      k--;
      y.first -= X;
      pq.push(y);
      zeros--;
    }
    while (zeros > 0 and k > 0) {
      y = pq.top();
      pq.pop();
      k--;
      y.first += X;
      pq.push(y);
      zeros--;
    }
  } else if (sgn > 0) {
    y = pq.top();
    pq.pop();
    int ch = min(k, abs(y.first) / X + 1);
    if (y.first > 0)
      y.first -= 1LL * ch * X;
    else
      y.first += 1LL * ch * X;
    k -= ch;
    pq.push(y);
  }
  while (k > 0) {
    y = pq.top();
    pq.pop();
    if (y.first > 0)
      y.first += X;
    else
      y.first -= X;
    pq.push(y);
    k--;
  }
  while (!pq.empty()) {
    a[pq.top().second] = pq.top().first;
    pq.pop();
  }
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << endl;
  return 0;
}
