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;
bool sa(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
int dif(string s, string ss) {
  int c = 0;
  for (int i = 0; i < 32; i++) {
    if (s[i] != ss[i]) {
      c++;
    }
  }
  return c;
}
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  int arr[m + 1];
  for (int i = 0; i < m + 1; i++) {
    int nn;
    cin >> nn;
    arr[i] = nn;
  }
  int c = 0;
  string s, ss;
  s = std::bitset<32>(arr[m]).to_string();
  for (int i = 0; i < m; i++) {
    ss = std::bitset<32>(arr[i]).to_string();
    int akmal = dif(s, ss);
    if (akmal <= k) {
      c++;
    }
  }
  cout << c << endl;
  return 0;
}