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;
int dx[]{1, -1, 0, 0, 1, -1, 1, -1};
int dy[]{0, 0, 1, -1, 1, -1, -1, 1};
long long suf(long long x) { return (x * (x + 1)) / 2; }
vector<vector<pair<int, long long> > > adj;
const long long mod = 1e9 + 7, size = 1000006, oi = INT_MAX, ol = INT64_MAX;
long long n, m, T, z, x, k, y;
int main() {
  while (cin >> n >> k >> x) {
    vector<pair<long long, int> > v(n);
    int si = 0;
    for (int i = 0; i < n; i++) {
      cin >> v[i].first;
      v[i].second = i;
      if (v[i].first < 0) si++;
    }
    sort(((v).begin()), ((v).end()));
    si %= 2;
    if (si == 0) {
      long long near = ol, id = 0;
      for (int i = 0; i < n; i++) {
        if (near > abs(v[i].first)) {
          near = abs(v[i].first), id = i;
        }
      }
      bool up = v[id].first < 0;
      while (k > 0 && ((v[id].first < 0) == up)) {
        if (up)
          v[id].first += x;
        else
          v[id].first -= x;
        k--;
      }
    }
    vector<bool> vl(n);
    for (int i = 0; i < n; i++)
      if (v[i].first < 0) vl[v[i].second] = 1, v[i].first *= -1;
    priority_queue<pair<long long, int>, vector<pair<long long, int> >,
                   greater<pair<long long, int> > >
        q;
    for (int i = 0; i < n; i++) {
      q.push(v[i]);
    }
    while (k > 0) {
      long long tp = q.top().first;
      int idx = q.top().second;
      q.pop();
      tp += x;
      q.push(make_pair(tp, idx));
      k--;
    }
    vector<long long> pr(n);
    while (!q.empty())
      pr[q.top().second] = q.top().first * (vl[q.top().second] ? -1 : 1),
      q.pop();
    for (int i = 0; i < n; i++) cout << pr[i] << " ";
    cout << endl;
  }
  return 0;
}
