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;
vector<int> sizes;
long long v[200005];
vector<pair<int, int> > p;
int main() {
  int n, k, x;
  cin >> n >> k >> x;
  int sign = 0;
  set<pair<long long, int> > neg;
  set<pair<long long, int> > pos;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    if (v[i] < 0) {
      sign ^= 1;
      neg.insert(make_pair(v[i], i));
    } else {
      pos.insert(make_pair(v[i], i));
    }
  }
  if (!sign) {
    bool from_a = false;
    if (neg.empty()) {
      from_a = true;
    }
    if (!pos.empty() && !neg.empty()) {
      long long a = pos.begin()->first;
      long long b = neg.rbegin()->first;
      b = -b;
      if (b > a) {
        from_a = true;
      }
    }
    while (!sign && k) {
      if (from_a) {
        pair<long long, int> p = *pos.begin();
        pos.erase(p);
        p.first -= x;
        if (p.first >= 0)
          pos.insert(p);
        else {
          neg.insert(p);
          sign ^= 1;
        }
      } else {
        pair<long long, int> p = *neg.rbegin();
        neg.erase(p);
        p.first += x;
        if (p.first <= 0)
          neg.insert(p);
        else {
          pos.insert(p);
          sign ^= 1;
        }
      }
      k--;
    }
  }
  while (k--) {
    if (!pos.empty() && neg.empty()) {
      pair<long long, int> p = *pos.begin();
      pos.erase(p);
      p.first += x;
      pos.insert(p);
    }
    if (pos.empty() && !neg.empty()) {
      pair<long long, int> p = *neg.rbegin();
      neg.erase(p);
      p.first -= x;
      neg.insert(p);
    }
    if (!pos.empty() && !neg.empty()) {
      pair<long long, int> p = *pos.begin();
      pair<long long, int> n = *neg.rbegin();
      long long a = p.first;
      long long b = -n.first;
      if (a > b) {
        neg.erase(n);
        n.first -= x;
        neg.insert(n);
      } else {
        pos.erase(p);
        p.first += x;
        pos.insert(p);
      }
    }
  }
  while (!pos.empty()) {
    pair<long long, int> x = *pos.begin();
    v[x.second] = x.first;
    pos.erase(x);
  }
  while (!neg.empty()) {
    pair<long long, int> x = *neg.begin();
    v[x.second] = x.first;
    neg.erase(x);
  }
  bool firstT = true;
  for (int i = 0; i < n; i++) {
    if (firstT)
      firstT = false;
    else
      cout << " ";
    cout << v[i];
  }
  cout << endl;
  return 0;
}
