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.io.PrintWriter;
import java.util.StringTokenizer;

public class B {
  FastScanner in;
  PrintWriter out;

  public void run() {
    try {
      in = new FastScanner(new InputStreamReader(System.in));
      out = new PrintWriter(System.out);
      solve();
      out.close();
    } catch (IOException e) {
      e.printStackTrace();
    }
  }

  public void solve() throws IOException {
    int n = in.nextInt();
    int m = in.nextInt();
    int k = in.nextInt();
    
    int c = 0;
    int [] a = new int[m];
    int fedya = 0;
    for (int i=0; i<m; i++) {
      a[i] = in.nextInt();
    }
    fedya = in.nextInt();
    
    int f = 0;
    int res = 0;
    int localCounter = 0;
    int counter = 0;
    StringBuilder sb = new StringBuilder();
    for (int i=0; i<m; i++) {
      f = fedya;
      res = f ^ a[i];
      localCounter = 0;
      sb = new StringBuilder();
      while (res > 0) {
        if ((res & 1) > 0) {
          localCounter++;
          sb.append(1);
        } else {
          sb.append(0);
        }
        sb.append(' ');
        res >>= 1;
      }
      //System.out.println(sb);
      if (localCounter <= k) {
        counter++;
      }
    }
    out.println(counter);
  }

  class FastScanner {
    BufferedReader br;
    StringTokenizer st;

    FastScanner(InputStreamReader in) {
      br = new BufferedReader(in);
    }

    String nextLine() {
      String str = null;
      try {
        str = br.readLine();
      } catch (IOException e) {
        e.printStackTrace();
      }

      return str;
    }

    String next() {
      while (st == null || !st.hasMoreTokens()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
          e.printStackTrace();
        }
      }
      return st.nextToken();
    }

    int nextInt() {
      return Integer.parseInt(next());
    }
  }

  public static void main(String[] arg) {
    B o = new B();
    o.run();
  }

}