Problem:
Search II

You are given a sequence of n integers S and a sequence of different q integers T. Write a program which outputs C, the number of integers in T which are also in the set S.

Input

In the first line n is given. In the second line, n integers are given. In the third line q is given. Then, in the fourth line, q integers are given.

Output

Print C in a line.

Constraints

Elements in S is sorted in ascending order

n ≤ 100000

q ≤ 50000

0 ≤ an element in S ≤ 109

0 ≤ an element in T ≤ 109

Sample Input 1

5
1 2 3 4 5
3
3 4 1

Sample Output 1

3

Sample Input 2

3
1 2 3
1
5

Sample Output 2

0

Sample Input 3

5
1 1 2 2 3
2
1 2

Sample Output 3

2

Notes

Solution in JAVA (judge verdict: Accepted)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author silviase
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        binarySearch solver = new binarySearch();
        solver.solve(1, in, out);
        out.close();
    }

    static class binarySearch {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = in.nextInt();
            int q = in.nextInt();
            int cnt = 0;
            for (int i = 0; i < q; i++) {
                int key = in.nextInt();
                int ng = -1;
                int ok = a.length;
                while (Math.abs(ok - ng) > 1) {
                    int mid = (ok + ng) / 2;
                    if (a[mid] >= key) {
                        ok = mid;
                    } else {
                        ng = mid;
                    }
                }
                if (a[ok] == key) cnt++;
            }
            out.println(cnt);
        }

    }
}