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: Wrong Answer)

#include<stdio.h>
int input(int*[]);
int isMatch(int*[], int*[], int, int);
int binarySearch(int*, int*[], int);

main(){
  int n, q;
  int *S[100000], *T[50000];

  n = input(S);
  q = input(T);
 
  printf("%d\n", isMatch(S, T, n, q));
  return 0;
}

int input(int* array[]){
  int i, e;
  scanf("%d", &e);
  for(i = 0; i < e; i++){
    scanf("%d", &array[i]);
  } 
  return e;
}

int isMatch(int* S[], int* T[], int n, int q){
  int count = 0, i;

  for(i = 0; i < q; i++){
    count += binarySearch(T[i], S, n);
  }
  return count;
}

int binarySearch(int* key, int* S[], int right){
  int mid, left;
  left = 0;
  while(left < right){
    mid = (right + left) / 2;
    if(key == S[mid]){
      return 1;
    } else if(key > S[mid]){
      left = mid + 1;
    } else {
      right = mid;
    }
  }
  return 0;
}