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)

#include <iostream>
#include <vector>
#include <stack>
#include <queue>
#include <list>
#include <functional>
#include <algorithm>
#include <complex>
#include <map>
#include <cmath>
#include <string>
#include <cstdio>
#include <stdio.h>
#include <cstdlib>
#include <istream>
//------------------------------------define------------------------//
#define PB push_buck
#define MP make_pair
//------------------------------------typedef-----------------------//
typedef long long LL;

using namespace std;

int a, b, c, d, e, f, g, h, i, j, k, l, n;
vector<int> s1;
vector<pair<int,int> > s2;
stack<int> s3;
queue<int> s4;
list<int> s5;

//--------------------------------------------------------------//
int binafySearch(int key) {
	int left = 0;
	int right = n;
	int midle;
	int A[1000000];
	while (left < right) {
		midle = (left + right) / 2;
		if (key == A[midle]) {
			return 1;
		}
		else if (key < A[midle]) {
			right = midle;
		}else if (key > A[midle]) {
			left = midle;
		}
	}
}


int main() {
	int sum = 0;
	int A[1000000];
	int q;
	int key;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> A[i];
	}
	cin >> q;
	for (int i = 0; i < q; i++) {
		cin >> key;

		if (binafySearch(key)) {
			sum++;
		}
	}
	printf("%d", sum);
}