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 vocala(char c) {
  if (c == 'a' || c == 'A' || c == 'e' || c == 'E' || c == 'i' || c == 'I' ||
      c == 'o' || c == 'O' || c == 'u' || c == 'U')
    return true;
  if (c == 'Y' || c == 'y') return true;
  return false;
}
int isprime(long long n) {
  if (n == 1) return 0;
  if (n == 2) return 1;
  if (n % 2 == 0) return 0;
  for (int i = 3; i * i <= n; i += 2)
    if (n % i == 0) return 0;
  return 1;
}
int isfibo(long long n) {
  long long a = 5 * n * n + 4;
  long long b = a - 8;
  if (sqrt(a) == int(sqrt(a)) || sqrt(b) == int(sqrt(b))) return 1;
  return 0;
}
int gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
ifstream in("txt.in");
ofstream out("txt.out");
long long const nrmax = 1e18;
int v[1005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, k, ans = 0;
  cin >> n >> m >> k;
  for (int i = 1; i <= m + 1; ++i) {
    cin >> v[i];
  }
  for (int i = 1; i <= m; ++i) {
    long long x = v[i], y = v[m + 1], a = 0;
    while (x > 0 || y > 0) {
      if (x % 2 != y % 2) ++a;
      x /= 2;
      y /= 2;
    }
    if (a <= k) ++ans;
  }
  cout << ans << '\n';
  in.close();
  out.close();
  return 0;
}