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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const long long mod = 1e9 + 7;
const long long INF = 1e18 + 1LL;
const int inf = 1e9 + 1e8;
const double PI = acos(-1.0);
const int N = 2e5 + 100;
struct Node {
  long long v, fv, id;
  void myabs() { fv = abs(v); }
  bool friend operator<(Node a, Node b) { return a.fv > b.fv; }
} a;
priority_queue<Node> Q;
long long n, f, x, k, ans[N];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k >> x;
  for (int i = 1; i <= n; ++i) {
    cin >> a.v;
    a.id = i, a.myabs();
    Q.push(a);
    if (a.v < 0) f++;
  }
  if (f & 1) {
    while (k--) {
      Node now = Q.top();
      Q.pop();
      if (now.v < 0)
        now.v -= x;
      else
        now.v += x;
      now.myabs();
      Q.push(now);
    }
  } else {
    Node now = Q.top();
    Q.pop();
    while (k--) {
      int f1 = now.v >= 0 ? 1 : -1;
      if (now.v < 0)
        now.v += x;
      else
        now.v -= x;
      now.myabs();
      int f2 = now.v >= 0 ? 1 : -1;
      if (f1 + f2 == 0) {
        Q.push(now);
        break;
      }
    }
    if (k == -1) {
      Q.push(now);
      k = 0;
    }
    while (k--) {
      now = Q.top();
      Q.pop();
      if (now.v < 0)
        now.v -= x;
      else
        now.v += x;
      now.myabs();
      Q.push(now);
    }
  }
  int t = Q.size();
  while (!Q.empty()) {
    ans[Q.top().id] = Q.top().v, Q.pop();
  }
  for (int i = 1; i <= t; ++i) {
    cout << ans[i] << " ";
  }
  return 0;
}
