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.

Notes

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

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.

Examples

Input

5
1 2 3 4 5
3
3 4 1


Output

3


Input

3
1 2 3
1
5


Output

0


Input

5
1 1 2 2 3
2
1 2


Output

2
import java.util.*;

public class Main{
	public static void main(String args[]){
		Scanner sc = new Scanner(System.in);
		int n, q, cnt = 0, isExist;
		int[] S = new int[100000], T = new int[50000];
		
		n = sc.nextInt();
		for(int i = 0; i < n; i++){
			S[i] = sc.nextInt();
		}
		
		q = sc.nextInt();
		for(int i = 0; i < q; i++){
			T[i] = sc.nextInt();
		}
		
		for(int i = 0; i < q; i++){
			isExist = binarySearch(T[i], S, n);
			if(isExist == 1){
				cnt ++;
			}
		}
		
		System.out.println(cnt);
	}
	
	public static int binarySearch(int t, int[] S, int len){
		int first = 0, last = len - 1, middle;
		int isExist = 0;
		
		while(isExist == 0 && first <= last){
			middle = (first + last) / 2;
			if(S[middle] == t){
				isExist = 1;
			}
			else if(t < S[middle]){
				last = middle - 1;
			}
			else if(S[middle] < t){
				first = middle + 1;
			}
		}
		
		return isExist;
	}
}