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;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (T i : v) os << i << ", ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const set<T> &v) {
  for (T i : v) os << i << " ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const multiset<T> &v) {
  for (T i : v) os << i << " ";
  return os;
}
template <class T, class second>
ostream &operator<<(ostream &os, const pair<T, second> &v) {
  os << v.first << ' ' << v.second;
  return os;
}
template <class T, class second>
ostream &operator<<(ostream &os, const map<T, second> &v) {
  for (auto i : v) os << '(' << i.first << " => " << i.second << ')' << ' ';
  return os;
}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
int n, k;
long long x;
vector<long long> v;
void solve() {
  int zero = 0, pos = 0, neg = 0;
  for (long long &i : v) {
    if (i == 0)
      zero++;
    else if (i > 0)
      pos++;
    else
      neg++;
  }
  if (zero > k) {
    for (long long &i : v) cout << i << ' ';
    cout << "\n";
    return;
  }
  if (zero) {
    if (neg % 2 == 1) {
      for (long long &i : v)
        if (i == 0 && k > 0) i += x, k--;
    } else {
      bool f = 0;
      for (long long &i : v) {
        if (i == 0 && k > 0) {
          if (f == 0)
            i -= x;
          else
            i += x;
          f = 1;
          k--;
        }
      }
    }
  }
  int f = 0;
  for (long long &i : v)
    if (i < 0) f ^= 1;
  if (f == 0) {
    int idx = 0;
    for (int i = 0; i < n; i++) {
      if (abs(v[i]) < abs(v[idx])) idx = i;
    }
    if (v[idx] < 0) {
      while (v[idx] <= 0 && k > 0) {
        v[idx] += x;
        k--;
      }
    } else {
      while (v[idx] >= 0 && k > 0) {
        v[idx] -= x;
        k--;
      }
    }
  }
  set<pair<long long, int>> s;
  for (int i = 0; i < n; i++) {
    s.insert({abs(v[i]), i});
  }
  while (k > 0) {
    auto it = *s.begin();
    s.erase(s.begin());
    if (v[it.second] < 0)
      v[it.second] -= x;
    else
      v[it.second] += x;
    s.insert({abs(v[it.second]), it.second});
    k--;
  }
  for (long long &i : v) cout << i << ' ';
  cout << "\n";
}
int main() {
  fast();
  cin >> n >> k >> x;
  for (int i = 0; i < n; i++) {
    long long a;
    cin >> a;
    v.push_back(a);
  }
  solve();
}
