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 JAVA):

Solution:
import java.util.*;
import java.lang.*;
import java.io.*;

public class Main
{
    static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
    static int MOD=1000000000+7;
    static long factorial(long n){
        if(n==1) return 1;
        if(n==2) return 2;
        if(n==3) return 6;
        return n*factorial(n-1);
    }
    static long gcd(long A,long B){
        if(B==0) return A;
        return gcd(B,A%B);
    }
    
    static long fast_expo(long x,long n){
        if(n==0) return 1;
        if((n&1)==0) return fast_expo((x*x)%MOD,n/2);
        return ((x%MOD)*fast_expo((x*x)%MOD,(n-1)/2));
    }
    
    static boolean isPrime(long n){
        if(n<=1) return false;
        if(n<=3) return true;
        if(n%2==0 || n%3==0) return false;
        for(int i=5;i*i<=n;i+=6)
            if(n%i==0 || n%(i+2)==0) return false;
        return true;
    }
    public static void main (String[] args) throws java.lang.Exception
    {
        FastReader sc=new FastReader();
        int n=sc.nextInt(),m=sc.nextInt(),k=sc.nextInt();
        int a[]=new int[m];
        for(int i=0;i<m;i++) a[i]=sc.nextInt();
        int p=sc.nextInt();
        int count=0;
        for(int i=0;i<m;i++){
        	int tmp=k,x=a[i]^p;
        	while(x>0){
        		x&=(x-1);
        		tmp--;
        	}
        	if(tmp>=0) count+=1;
        }
        System.out.println(count);
    }
}