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 ceill(long long a, long long b) { return a / b + bool(a % b); }
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    x = (x * x) % 1000000007;
    y >>= 1;
  }
  return res;
}
void bin(unsigned n, string &str) {
  str = "";
  if (n > 1) bin(n >> 1, str);
  str += to_string(n & 1);
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
void primeFactors(long long n, vector<long long> &vec) {
  while (n % 2 == 0) vec.push_back(2), n /= 2;
  for (long long i = 3; i <= sqrt(n); i += 2)
    while (n % i == 0) vec.push_back(i), n /= i;
  if (n > 2) vec.push_back(n);
}
void solve() {
  long long n, m, k;
  cin >> n >> m >> k;
  long long arr[m + 1];
  for (long long i = 0; i < m + 1; i++) cin >> arr[i];
  long long tmp = arr[m];
  long long cnt = 0;
  for (long long i = 0; i < m; i++) {
    long long t = tmp ^ arr[i];
    if (__builtin_popcountll(t) <= k) cnt++;
  }
  cout << cnt << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long tc = 1;
  for (long long TTT = 1; TTT <= tc; TTT++) {
    solve();
  }
}