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;
void file() {}
const int MN = 1e4 + 1e5;
const int MW = 1e3 + 5;
const int OO = 1e9 + 5;
int main() {
  file();
  int n, m, k, p[1002]{0}, ans = 0;
  scanf("%d", &n), scanf("%d", &m), scanf("%d", &k);
  for (int i = 0; i <= m and scanf("%d", &p[i]); ++i)
    ;
  for (int i = 0, df = 0; i < m; ans += (df <= k), df = 0, ++i)
    for (int j = 0; j < 21 and df <= k; ++j)
      if ((p[m] & (1 << j)) != (p[i] & (1 << j))) df++;
  printf("%d", ans);
  return 0;
}