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)

/*
 * Binary Search.c
 *
 *  Created on: 2015/02/15
 */

#include <stdio.h>

int S[100000]={0},T[100000]={0};
int flag[1000]={0};

void printarray(int a[],int n){
	int i;
	for(i=0;i<n;i++){
		printf("%d ",a[i]);
	}printf("\n");
}

int main(){
	int n,q;
	int i,j,counter=0;
	int start,end;

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

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


	start=0,end=n-1;

	for(i=0;i<q;i++){
		while(end - start >= 2){
			if(S[(start+end)/2] > T[i]){
				end = (start+end)/2;
			}
			if(S[(start+end)/2] < T[i]){
				start = (start+end)/2;
			}
		}

		if(S[start] == T[i]) flag[i] = 1;
		if(S[end] == T[i]) flag[i] = 1;
	}

	for(i=0;i<q;i++){
		if(flag[i]==1)counter++;
	}
	printf("%d\n",counter);

	return 0;
}