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;
const int N = 200010;
int n, k;
long long x, seq[N];
struct Node {
  long long x;
  int id;
  bool operator<(const Node &rhs) const { return abs(x) > abs(rhs.x); }
};
priority_queue<Node> que;
vector<Node> ans;
bool cmp(const Node &lhs, const Node &rhs) { return lhs.id < rhs.id; }
int main() {
  ios::sync_with_stdio(false);
  while (!que.empty()) que.pop();
  ans.clear();
  cin >> n >> k >> x;
  int zcnt = 0;
  for (int i = 0; i < n; ++i) {
    cin >> seq[i];
    que.push((Node){seq[i], i});
    if (seq[i] < 0) ++zcnt;
  }
  while (k) {
    Node tp = que.top();
    que.pop();
    if (zcnt & 1) {
      if (tp.x >= 0)
        tp.x += x;
      else
        tp.x -= x;
    } else {
      if (tp.x < 0) {
        tp.x += x;
        if (tp.x >= 0) zcnt--;
      } else {
        tp.x -= x;
        if (tp.x < 0) zcnt++;
      }
    }
    que.push(tp);
    --k;
  }
  while (!que.empty()) {
    ans.push_back(que.top());
    que.pop();
  }
  sort(ans.begin(), ans.end(), cmp);
  for (int i = 0; i < ans.size(); ++i) {
    cout << ans[i].x;
    if (i < ans.size() - 1)
      cout << " ";
    else
      cout << endl;
  }
  return 0;
}
