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 int mulmod(long long int a, long long int b, long long int m) {
  long long int q =
      (long long int)(((long double)a * (long double)b) / (long double)m);
  long long int r = a * b - q * m;
  if (r > m) r %= m;
  if (r < 0) r += m;
  return r;
}
template <typename T>
T MOD(T a, T b) {
  return (a < b ? a : a % b);
}
template <typename T>
T expo(T e, T n) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = x * p;
    p = p * p;
    n >>= 1;
  }
  return x;
}
template <typename T>
T power(T e, T n, T m) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = mulmod(x, p, m);
    p = mulmod(p, p, m);
    n >>= 1;
  }
  return x;
}
template <typename T>
T gcd(T a, T b) {
  if (a % b) return gcd(b, a % b);
  return b;
}
template <typename T>
T lcm(T a, T b) {
  return (a * (b / gcd(a, b)));
}
set<pair<long long int, long long int> > s;
long long int a[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, k, x;
  cin >> n >> k >> x;
  bool neg = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    s.insert({abs(a[i]), i});
    neg ^= (a[i] < 0);
  }
  while (k) {
    pair<long long int, long long int> p = *s.begin();
    if (neg) {
      s.erase(s.begin());
      s.insert({p.first + x, p.second});
      if (a[p.second] < 0)
        a[p.second] -= x;
      else
        a[p.second] += x;
      --k;
    } else {
      s.erase(s.begin());
      if (a[p.second] < 0) {
        while (a[p.second] <= 0 and k) a[p.second] += x, --k;
        if (a[p.second] > 0) neg = 1;
      } else {
        while (a[p.second] >= 0 and k) a[p.second] -= x, --k;
        if (a[p.second] < 0) neg = 1;
      }
      s.insert({abs(a[p.second]), p.second});
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << a[i] << " ";
  }
  return 0;
}
