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 n, m, k, i = 0, data, compara, cant, cont0 = 0, amigo = 0;
int bit(int n) {
  int cont = 0;
  while (n != 0) {
    n = n / 2;
    cont++;
  }
  return cont;
}
int main() {
  vector<int> datos;
  cin >> n >> m >> k;
  n = pow(2, n);
  while (i <= m) {
    cin >> data;
    datos.push_back(data);
    i++;
  }
  i = 0;
  while (i < m) {
    if (datos[i] < n) {
      cont0 = 0;
      if (datos[i] > datos[m]) {
        cant = bit(datos[i]);
        compara = ((datos[i]) ^ (datos[m]));
        for (int i = 0; i < cant; i++) {
          if (compara & (1 << i)) cont0++;
        }
      } else {
        cant = bit(datos[m]);
        compara = ((datos[i]) ^ (datos[m]));
        for (int i = 0; i < cant; i++) {
          if (compara & (1 << i)) cont0++;
        }
      }
    }
    if (cont0 <= k) amigo++;
    i++;
  }
  cout << amigo << endl;
}