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;
long long modpow(long long a, long long b, long long MOD) {
  long long res = 1;
  a %= MOD;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
  }
  return res;
}
template <typename T>
static T mod(T a, T m) {
  a %= m;
  if (a < 0) a += m;
  return a;
}
template <typename T>
static T inv(T a, T m) {
  a = mod(a, m);
  if (a <= 1) return a;
  return mod((1 - inverse(m % a, a) * m) / a, m);
}
struct hashp {
  template <class T1, class T2>
  std::size_t operator()(const std::pair<T1, T2> &p) const {
    auto h1 = std::hash<T1>{}(p.first);
    auto h2 = std::hash<T2>{}(p.second);
    return h1 ^ h2;
  }
};
long long i = 0, j = 0;
long long M = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m, k;
  cin >> n >> m >> k;
  vector<long long> a;
  long long f, x;
  for (i = 0; i < m; i++) {
    cin >> x;
    a.push_back(x);
  }
  cin >> f;
  long long ans = 0;
  for (i = 0; i < m; i++) {
    x = f ^ a[i];
    if (__builtin_popcountll(x) <= k) ans++;
  }
  cout << ans << '\n';
  return 0;
}