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 ans = 1, p, t, n, k, m, a[1001], flag;
bool b[1001][21];
int doo(int h) {
  for (int i = 0; i < h; ++i) ans *= 2;
  t = ans;
  ans = 1;
  return t;
}
signed main() {
  cin >> n >> m >> k;
  m++;
  for (int i = 0; i < m; ++i) {
    cin >> a[i];
    a[i]++;
    for (int j = 0; j < n; ++j)
      a[i] > doo(n - j - 1) ? b[i][j] = 1, a[i] -= doo(n - j - 1) : b[i][j] = 0;
  }
  m--;
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j < n; ++j) {
      b[i][j] != b[m][j] ? p++ : p = p;
    }
    p <= k ? flag++ : flag = flag;
    p = 0;
  }
  cout << flag << endl;
  return 0;
}