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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class CF{public static void main(String[]args)throws IOException{S s=new S();s.s();s.output();}}class S{
    void s() throws IOException {
        int n = NI();
        int m = NI();
        int k = NI();
        int[] a = new int[m];
        inInt(a, m);
        int t = NI();
        int res = 0;
        for (int i = 0; i < m; i++)
            if (calc(a[i] ^ t) <= k)
                res++;
        sout(res);
    }

    int calc(int x) {
        int res = 0;
        while (x > 0) {
            res += x % 2;
            x /= 2;
        }
        return res;
    }

    BufferedReader stdin=new BufferedReader(new InputStreamReader(System.in));StringTokenizer st=new StringTokenizer("");long LINF=Long.MAX_VALUE;
    StringBuilder output=new StringBuilder();final int INF=Integer.MAX_VALUE;void sout(Object x){output.append(x.toString()).append('\n');}
    void inLong(long[]a,int n){for(int i=0;i<n;i++)a[i]=NL();}int min(int i1,int i2){return i1<i2?i1:i2;}double ND(){return Double.parseDouble(NS());}
    long min(long i1,long i2){return i1<i2?i1:i2;}int max(int i1,int i2){return i1>i2?i1:i2;}long max(long i1,long i2){return i1>i2?i1:i2;}
    String NS(){while(!st.hasMoreTokens())try{st=new StringTokenizer(stdin.readLine());}catch(IOException ignored){}return st.nextToken();}
    int NI(){return Integer.parseInt(NS());}long NL(){return Long.parseLong(NS());}String NLn()throws IOException{return stdin.readLine();}
    int abs(int x){return x<0?-x:x;}long abs(long x){return x<0?-x:x;}void sout(){output.append('\n');}void out(Object x){output.append(x.toString());}
    int mod(int x,int mod){return(x+mod)%mod;}void output(){System.out.print(output);}void inInt(int[]a,int n){for(int i=0;i<n;i++)a[i]=NI();}
}