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 M = 1e9 + 7;
long long mpe(long long base, long long exponent, long long modulus) {
  long long result = 1;
  while (exponent > 0) {
    if (exponent % 2 == 1) result = result * base % modulus;
    exponent = exponent >> 1;
    base = base * base % modulus;
  }
  return result;
}
long long arr[200005], sign[200005], mi = 1e16, po;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  ;
  long long n, k, x;
  cin >> n >> k >> x;
  long long f = 0;
  for (long long i = 0; i < n; ++i) {
    cin >> arr[i];
    if (arr[i] < 0) {
      f = 1 - f;
      sign[i] = 1;
    }
    if (abs(arr[i]) < mi) {
      mi = abs(arr[i]);
      po = i;
    }
  }
  long long tk = floor(mi * 1.0 / x) + 1;
  if (f == 0) {
    if (tk > k) {
      long long tmp = mi - k * x;
      arr[po] = tmp;
      for (long long i = 0; i < n; ++i) {
        tmp = abs(arr[i]);
        tmp = (sign[i] == 1) ? (tmp * -1) : tmp;
        cout << tmp << " ";
      }
      return 0;
    }
    k -= tk;
    sign[po] = 1 - sign[po];
    arr[po] = abs(mi - tk * x);
  }
  set<pair<long long, long long> > st;
  for (long long i = 0; i < n; ++i) {
    st.insert(make_pair(abs(arr[i]), i));
  }
  while (k--) {
    pair<long long, long long> tmp = (*st.begin());
    st.erase(tmp);
    tmp.first += x;
    st.insert(tmp);
  }
  for (auto i : st) {
    arr[i.second] = i.first;
  }
  long long tmp;
  for (long long i = 0; i < n; ++i) {
    tmp = abs(arr[i]);
    tmp = (sign[i] == 1) ? tmp * -1 : tmp;
    cout << tmp << " ";
  }
  return 0;
}
