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.String;
import java.lang.Math;
import java.util.Arrays;

public class Capitalization2
{
   
    
    public static void main(String[] args)
    {
Scanner sc=new Scanner(System.in);
int n,m,l;
n=sc.nextInt();
m=sc.nextInt();
l=sc.nextInt();
int sum=0;
int [] binary=new int[n+1];
int [] a=new int[m+1];

for(int i=0;i<=m;i++)
{
    a[i]=sc.nextInt();
    if(i==m)
    {
        int remsum=a[i];
        
     for(int k=n-1;k>=0;k--)
     {
         
         double res=Math.pow(2,k);
         int res1=(int)res;
         if(res>remsum)
         binary[k]=0;
         else
         {
             binary[k]=1;
             remsum=remsum-res1;
         }
         
         
         
     }
    }
}


for(int i=0;i<m;i++)
{
    int c=0;
    int remsum=a[i];
     int [] b=new int[n+1];   
     for(int k=n-1;k>=0;k--)
     {
         
         double res=Math.pow(2,k);
         int res1=(int)res;
       
         if(res1>remsum)
         b[k]=0;
         else
         {
             b[k]=1;
             remsum=remsum-res1;
         }
         if(b[k]!=binary[k])
         {
             c++;
         }
       
         
}
if(c<=l)
{
sum++;
c=0;
}

}
System.out.print(sum);

}
}