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;
vector<int> bin(int num, int typs) {
  vector<int> v;
  while (num > 0) {
    v.push_back(num % 2);
    num /= 2;
  }
  for (int i = v.size(); i < typs; i++) {
    v.push_back(0);
  }
  return v;
}
int main() {
  int typs, plyrs, diff, cnt = 0, diffbit = 0;
  cin >> typs >> plyrs >> diff;
  int army;
  vector<vector<int> > v;
  for (int i = 0; i <= plyrs; i++) {
    cin >> army;
    v.push_back(bin(army, typs));
  }
  for (int i = 0; i < v.size() - 1; i++) {
    diffbit = 0;
    for (int j = 0; j < v[i].size(); j++) {
      if (v[i][j] != v[v.size() - 1][j]) diffbit++;
    }
    if (diffbit <= diff) cnt++;
  }
  cout << cnt;
  return 0;
}