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 calcpow(float gr, long long n) {
  long long ans = 1.0;
  while (n != 0) {
    if (n % 2 == 1) {
      ans = ans * gr;
    }
    gr = gr * gr;
    n /= 2;
  }
  return ans;
}
int noofdiffbits(int n) {
  long long c = 0;
  while (n != 0) {
    n = n & (n - 1);
    c++;
  }
  return c;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, k;
  cin >> n >> m >> k;
  long long a[m + 1];
  for (int i = 0; i < m + 1; ++i) {
    cin >> a[i];
  }
  long long c = 0;
  for (int i = 0; i < m; ++i) {
    int l = a[m] ^ a[i];
    if (noofdiffbits(l) <= k) {
      c++;
    }
  }
  cout << c;
}