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")
using namespace std;
const int INF = signed(1e9) + 7;
const long double EPS = 1e-7, Pi = acos(-1);
bitset<21> bts[1001];
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < m + 1; ++i) {
    int q;
    cin >> q;
    bts[i] |= q;
  }
  int ans = 0;
  for (int i = 0; i < m; ++i) {
    ans += int((bts[i] ^ bts[m]).count() <= k);
  }
  cout << ans << '\n';
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(10);
  int tests = 1;
  while (tests--) {
    solve();
  }
}