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 <stdio.h>
#include <stdlib.h>

int binary_search(int *S, int st, int ed, int x);

int main(int argc, char *argv[])
{
  int n, q, i, j;
  int *S, *T;
  int found;

  scanf("%d", &n);
  S = (int *)malloc(sizeof(int) * n);
  for (i = 0; i < n; i++) scanf("%d", &S[i]);

  scanf("%d", &q);
  T = (int *)malloc(sizeof(int) * q);
  for (i = 0; i < q; i++) scanf("%d", &T[i]);

  found = 0;
  for (i = 0; i < q; i++) found += binary_search(S, 0, n - 1, T[i]);

  printf("%d\n", found);

  free(S);
  free(T);

  return 0;
}

int binary_search(int *S, int st, int ed, int x)
{
  int i;

  if (ed - st < 20) {
    for (i = st; i <= ed; i++) {
      if (S[i] == x) return 1;
    }
    return 0;
  } else {
    if (S[(ed-st) / 2] > x) return binary_search(S, st, (ed-st) / 2, x);
    else if (S[(ed-st) / 2] < x) return binary_search(S, (ed-st) / 2, ed, x);
    else return 1;
  }
}