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 C (judge verdict: Time Limit Exceeded)

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>


#define MAX 100000


int kensaku(int S[], int T[], int Q, int N)
{

	int i = 0, j = 0, k = 0, z = 0;
	int zcount = 0,EOFcount = 0, V[MAX];

	for (j = 0; j < N; j++) {
		for (i = 0; i < Q; i++) {
			if (T[i] == S[j]) {//TとSが同じ値なら
				for (k = 0; k < zcount; k++) {
					if (T[i] == V[k]) {//TとVが同じ値なら
						EOFcount++;
						break;
					}
				}
					if (EOFcount == 0) {
						V[z] = T[i];
						z++;
						zcount++;
						}
				EOFcount = 0;
			}//if文ここまで
		}
	}

	return zcount;
}



int main(void)
{
	int i;
	int n, q, count = 0;
	int s[MAX], t[MAX];

	/////一つ目の配列の要素数と要素/////
	scanf("%d", &n);

	for (i = 0; i < n; i++) {
		scanf("%d", &s[i]);
	}
	////////////////////////////////////

	/////二つ目の配列の要素数と要素/////
	scanf("%d", &q);

	for (i = 0; i < q; i++) {
		scanf("%d", &t[i]);
	}
	////////////////////////////////////

	printf("%d\n", count = kensaku(s, t, q, n));

	return 0;
}