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.io.*;
public class FedorAndNewGame
{
    public static void main(String args[])throws IOException
    {
        BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
        String str1[] = rd.readLine().split(" ");
        int n = Integer.parseInt(str1[0]);
        int m = Integer.parseInt(str1[1]);
        int k = Integer.parseInt(str1[2]);
        int arr[] = new int[m];
        String a[] = new String[m];
        int count = 0;
        for(int i = 0; i < m; i++)
            arr[i] = Integer.parseInt(rd.readLine());
        String fdobin = binary(Integer.parseInt(rd.readLine()));
        for(int i = 0; i < m; i++)
            {
                a[i] = binary(arr[i]);
            }    
        for(int i = 0; i < m; i++)
        {
            int d = differ(a[i],fdobin);
            if(d <= k)
                count++;
        }
        System.out.println(count);
    }
    private static String binary(int xarmy)
    {
        String s = "";
        while(xarmy != 0)
        {
            s = (xarmy%2) + s;
            xarmy = xarmy/2;
        }
        return s;
    }
    private static int differ(String x, String fd)
    {
        int dif = 0;
        int j = 0;
        int l1 = x.length() - 1;
        int l2 = fd.length() - 1;
        String max = l1>=l2?x:fd;
        while(l1 >= 0 && l2 >= 0)
        {
            char xr = x.charAt(l1);
            char fdr = fd.charAt(l2);
            if((xr != fdr) && ( xr == '0' || fdr == '0'))
                dif++;
            l1--;
            l2--;    
        }
        j = l1>=l2?l1:l2;
        while(j>=0)
        {
            if(max.charAt(j) == '1')
                dif++;
            j--;
        }   
        return dif;
    }
}