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 Main 
{
    public static void main(String []args)
  {
      Scanner input = new Scanner(System.in);
      int n = input.nextInt();
      int m = input.nextInt();
      int k = input.nextInt();
      int[] armies = new int[m+1];
      for( int i=0; i<=m; i++) 
          armies[i] = input.nextInt();
      System.out.println(f(n,m,k,armies));
  }   
    
    public static int f( int n, int m, int k, int[] armies)
  {
      int count = 0;
      int[][] armyInBin = new int[m+1][n];
      for( int i=0; i<=m; i++) 
            armyInBin[i] = binary(n,armies[i]);
      for( int i=0; i<=m-1; i++)
         {
            int difCount = 0;
            for( int j=0; j<=n-1; j++)
                  if(armyInBin[i][j]!=armyInBin[m][j]) difCount++;
               
            if(difCount<=k) count++; 
         }
         
      return count;
  }
    
    public static int[] binary( int n, int num)
  {
      int[] result = new int[n];
      for( int j=0; j<=n-1; j++)
          result[j] = 0;
      int i = 0;
      while( num!=0 )
         {
            result[i++] = num%2;
            num /= 2;
         }
      
      return result;
  } 
}