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.io.BufferedReader;
import java.io.InputStreamReader;

public class Main
{
	public static void main(String[] args)throws Exception
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n  = Integer.parseInt(br.readLine());
		String[] g  = br.readLine().split(" ");
		int [] ar = new int[n];
		for( int i = 0 ; i< n ; i++){
			ar[i] = Integer.parseInt(g[i]);
		}
		int m  = Integer.parseInt(br.readLine());
		int ans = 0;
		g  = br.readLine().split(" ");
		for(int i = 0 ; i <m; i++)
		{
			if(binarySearch(ar,Integer.parseInt(g[i])) != -1) ans++ ;	
		}
		System.out.println(ans);
	}
	public static int binarySearch(int[] c, int key)
	{
		int len = c.length;
		int left = 0 , right = len;
		int mid = (right + left) / 2 ;
		while(left < right){
			if(c[mid] == key) return mid;
			else if(c[mid] < key) left = mid+1;
			else { right = mid;  }
			mid = (right + left) / 2 ;
		}
		return -1;
	}
	
}