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>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
std::pair<int, int> DR[] = {{-1, 0}, {0, 1},   {1, 0}, {0, -1},
                            {-1, 1}, {-1, -1}, {1, 1}, {1, -1}};
using namespace std;
int cone(int n) {
  int cnt = 0;
  while (n) {
    n = (n - 1) & n;
    cnt++;
  }
  return cnt;
}
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  int b[m + 2];
  for (int i = 0; i <= m; i++) {
    cin >> b[i];
  }
  int cnt = 0;
  for (int j = 0; j < m; j++) {
    if (cone(b[m] ^ b[j]) <= k) cnt++;
  }
  cout << cnt << endl;
  return 0;
}