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;
string to_binary(int h) {
  string m;
  while (h != 0) {
    m += to_string(h % 2);
    h /= 2;
  }
  return m;
}
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  int ans = 0, nm;
  string o;
  string f;
  m++;
  vector<string> v;
  while (m--) {
    cin >> nm;
    if (m == 0) {
      f = to_binary(nm);
    } else {
      o = to_binary(nm);
      v.push_back(o);
    }
  }
  for (int i = 0; i < v.size(); i++) {
    int dif = 0;
    for (int j = 0; j < min(v[i].size(), f.size()); j++) {
      if (v[i][j] != f[j]) {
        dif++;
      }
    }
    if (v[i].size() < f.size()) {
      for (int j = v[i].size(); j < f.size(); j++) {
        if (f[j] == '1') {
          dif++;
        }
      }
    } else {
      for (int j = f.size(); j < v[i].size(); j++) {
        if (v[i][j] == '1') {
          dif++;
        }
      }
    }
    if (dif <= k) {
      ans++;
    }
  }
  cout << ans;
  return 0;
}