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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k, m, c = 0, flag, counter = 0, q = 0;
  long long int x;
  cin >> n >> m >> k;
  vector<vector<int> > v(m + 1);
  for (int i = 0; i <= m; ++i) {
    cin >> x;
    if (x == 0) {
      v[i].push_back(0);
      continue;
    } else {
      while (x > 0) {
        v[i].push_back(x % 2);
        x = x / 2;
      }
    }
  }
  for (int i = 0; i < m; ++i) {
    flag = 0;
    c = 0;
    q = 0;
    while (q < max(v[i].size(), v[m].size())) {
      if (q >= v[i].size() && q < v[m].size()) {
        if (v[m][q] == 1) c++;
      } else if (q < v[i].size() && q >= v[m].size()) {
        if (v[i][q] == 1) c++;
      } else if (q < v[i].size() && q < v[m].size()) {
        if ((v[i][q] == 1 && v[m][q] == 0) || (v[i][q] == 0 && v[m][q] == 1))
          c++;
      }
      if (c > k) {
        flag = 1;
        break;
      }
      q++;
    }
    if (flag == 1)
      continue;
    else
      counter++;
  }
  cout << counter;
  return 0;
}