Problem:
Prime Numbers

A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7.

Write a program which reads a list of N integers and prints the number of prime numbers in the list.

Input

The first line contains an integer N, the number of elements in the list.

N numbers are given in the following lines.

Output

Print the number of prime numbers in the given list.

Constraints

1 ≤ N ≤ 10000

2 ≤ an element of the list ≤ 108

Sample Input 1

5
2
3
4
5
6

Sample Output 1

3

Sample Input 2

11
7
8
9
10
11
12
13
14
15
16
17

Sample Output 2

4

Solution in C (judge verdict: Runtime Error)

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

int sieve(char *list, long max) {
  int i,j;
  long t;
  list[0] = 1;
  list[1] = 1;

  t = 1;
  for(i = 2; i <= (int)sqrt(max); i++) {
    if(list[i] == 0) {
      for(j = i; i*(long)j <= max; j++) {
	if(list[i*j] == 0) {
	  list[i*j] = 1;
	  t++;
	}
      }
    }
  }

  return (int)(max - t);
}

int prime_check(long n, long *list, int l) {
  int i,t;
  t = 0;
  for(i = 0; i < l && list[i] < n; i++) {
    if(n % list[i] == 0) {
      t = 1;
      break;
    }
  }
  if(n == 0 || n == 1) t = 1;
  
  return t;
}


int main() {
  int N,i,t,d;
  long li;
  scanf("%d", &N);
  long a[N],max;
  max = 0;
  for(i = 0; i < N; i++) {
    scanf("%ld", a+i);
    if(a[i] > max) max = a[i];
  }
  char *prime_list;
  prime_list = (char *)malloc(sizeof(char) * (max/2+1));
  memset(prime_list, 0, max/2+1);
  d = sieve(prime_list,max/2);
  
  long d_list[d];
  i = 0;
  for(li = 0; li <= max/2; li++) {
    if(prime_list[li] == 0) {
      d_list[i] = li;
      i++;
      if(i == d) break;
    }
  }
  free(prime_list);


  t = 0;
  for(i = 0; i < N; i++) {
    if(prime_check(a[i], d_list, d) == 0) t++;
  }
  printf("%d\n", t);
  
  return 0;
}