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.Scanner;

public class B_Fedor_and_New_Game {
	static Scanner input = new Scanner(System.in);

	public static void main(String[] args) {
		int n=input.nextInt();
		int m=input.nextInt();
		int k=input.nextInt();
		
		String[] arr=new String[m+1];
		
		for (int i = 0; i < m+1; i++) {
			arr[i]=Integer.toBinaryString(input.nextInt());
		}
		int count1=0;
		int size1=arr[m].length();
		for (int i = 0; i < m; i++) {
			int count=0;
			int comp=arr[i].length()<size1?arr[i].length():size1;
			int j=0;
			int n1=arr[i].length()-1;
			int n2=size1-1;
			int flag=0;
			
			while(comp-j-1>-1){
				if(arr[i].charAt(n1)!=arr[m].charAt(n2)){
					count++;
				}
				if(count>k)
					flag=1;
				j++;
				n1--;
				n2--;
				
			}
			
			String s=n2>n1?arr[m]:arr[i];
			n2=n2>n1?n2:n1;
			
			while(n2>-1){
				if(s.charAt(n2)!='0')
					count++;
				
				if(count>k)
					flag=1;
				
				n2--;
			}
			
			if(flag==0)
				count1++;
		}
		
		System.out.println(count1);
	}
}