B. Fedor and New Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ktypes 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.

Sample test(s)

input

7 3 1
8
5
111
17

output

0

input

3 3 3
1
2
3
4

output

3

解题说明：此题其实是求异或，然后统计1的个数即可。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
using namespace std;

int main()
{
  int n, m, k, p[1001], i, xxor, ct;
  int count = 0;
  scanf("%d %d %d", &n, &m, &k);
  for(i=0;i<=m;i++)
  {
	  scanf("%d", &p[i]);
  }
  for(i=0;i<m;i++)
  {
    ct = 0;
    xxor = p[m] ^ p[i];
    while(xxor)
    {
      ct++;
      xxor = xxor & (xxor-1);
    }
    if(ct<=k)
    {
		count++;
	}
  }
  printf("%d\n", count);
  return 0;
}