Problem: After you had helped George and Alex to move in the dorm, they went to help their friend Fedor play a new computer game «Call of Soldiers 3».

The game has (m + 1) players and n types of soldiers in total. Players «Call of Soldiers 3» are numbered form 1 to (m + 1). Types of soldiers are numbered from 0 to n - 1. Each player has an army. Army of the i-th player can be described by non-negative integer xi. Consider binary representation of xi: if the j-th bit of number xi equal to one, then the army of the i-th player has soldiers of the j-th type. 

Fedor is the (m + 1)-th player of the game. He assume that two players can become friends if their armies differ in at most k types of soldiers (in other words, binary representations of the corresponding numbers differ in at most k bits). Help Fedor and count how many players can become his friends.

Input

The first line contains three integers n, m, k (1 ≤ k ≤ n ≤ 20; 1 ≤ m ≤ 1000).

The i-th of the next (m + 1) lines contains a single integer xi (1 ≤ xi ≤ 2n - 1), that describes the i-th player's army. We remind you that Fedor is the (m + 1)-th player.

Output

Print a single integer — the number of Fedor's potential friends.

Examples

Input

7 3 1
8
5
111
17


Output

0


Input

3 3 3
1
2
3
4


Output

3

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long n, m, k;
    cin >> n >> m >> k;
    string a[m + 1];
    long long temp;
    for (long long int i = 0; i < m + 1; i++) {
      cin >> temp;
      long long cnt = 0;
      while (temp > 0) {
        if (temp % 2 == 0)
          a[i] = '0' + a[i];
        else
          a[i] = '1' + a[i];
        temp /= 2;
        cnt++;
      }
      for (long long int j = cnt + 1; j < n + 2; j++) a[i] = '0' + a[i];
    }
    long long ans = 0;
    for (long long int i = 0; i < m; i++) {
      long long cnt = 0;
      for (long long int j = 0; j < n + 2; j++)
        if (a[i][j] != a[m][j]) cnt++;
      if (cnt <= k) ans++;
    }
    cout << ans << '\n';
  }
  return 0;
}