codeforces—— 467B —— Fedor and New Game

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 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 18511117

output

0

input

3 3 31234

output

3

输入n,m,k.输入m+1个数，取值小于等于n，问最后输入的数与其余的数，在二进制下差别不超过k位的有多少个

#include<iostream>#include<cmath>#include<cstdio>#include<map>#include<set>#include<cstring>#include<string>#include<algorithm>#define min 1e-6using namespace std;int main(){    int temp,n,m;    while(cin>>temp>>n>>m)    {        int self,com[1008];        temp=n;        while(temp--)            cin>>com[temp];        cin>>self;        int ans=0;        for(int i=0; i<n; i++)        {            temp=m;            for(int j=0; j<32; j++)                if(((com[i]>>j)&1)!=((self>>j)&1))                    temp--;            if(temp>=0)                ans++;        }        cout<<ans<<endl;    }    return 0;}