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: Time Limit Exceeded)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class Main {
	public static int count = 0;
	
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		
		int n = Integer.parseInt(br.readLine());
		String[] sready = (br.readLine()).split(" ");
		int[] s = new int[n];
		for (int i = 0; i < n; i++)
			s[i] = Integer.parseInt(sready[i]);
		
		int q = Integer.parseInt(br.readLine());
		String[] tready = (br.readLine()).split(" ");
		int[] t = new int[q];
		for (int i = 0; i < q; i++)
			t[i] = Integer.parseInt(tready[i]);
		
		for (int i = 0; i < q; i++)
			doBinarySearch(s, t[i]);
		
		pw.println(count);
		pw.close();
	}
	
	public static void doBinarySearch(int[] s, int x) {
		int index1 = 0;
		int index2 = s.length - 1;
		if (x < s[index1])
			return;
		else if (x > s[index2])
			return;
		
		int point = (index2 - index1) / 2;
		do {
			if (s[point] == x) {
				count++;
				break;
			}
			else if (s[point] > x)
				index2 = point;
			else
				index1 = point;
			point = index1 + (index2 - index1) / 2;
		} while (index1 != index2);
	}
}