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 i, j, A[100001], m, k, n, w, t;
string l, r;
string Rec(long long e) {
  string s = "";
  while (e > 0) {
    s = s + char(e % 2 + 48);
    e = e / 2;
  }
  return s;
}
int main() {
  cin >> n >> m >> k;
  for (i = 1; i <= m + 1; i++) cin >> A[i];
  l = Rec(A[m + 1]);
  for (i = 1; i <= m; i++) {
    w = 0;
    r = Rec(A[i]);
    for (j = 0; j < max(r.size(), l.size()); j++) {
      if (j >= r.size() && j < l.size() && l[j] != '0')
        w++;
      else if (j < r.size() && j >= l.size() && r[j] != '0')
        w++;
      else if (j < r.size() && j < l.size() && r[j] != l[j])
        w++;
    }
    if (w <= k) t++;
  }
  cout << t;
}