Problem:
Score: 200 points

Problem Statement

The number 105 is quite special - it is odd but still it has eight divisors.
Now, your task is this: how many odd numbers with exactly eight positive divisors are there between 1 and N (inclusive)?

Constraints

N is an integer between 1 and 200 (inclusive).

Input

Input is given from Standard Input in the following format:

N

Output

Print the count.

Sample Input 1

105

Sample Output 1

1

Among the numbers between 1 and 105, the only number that is odd and has exactly eight divisors is 105.

Sample Input 2

7

Sample Output 2

0

1 has one divisor. 3, 5 and 7 are all prime and have two divisors. Thus, there is no number that satisfies the condition.

Solution in C (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
int main(){
  int N;
  scanf("%d", N);
  int count = 0;
  for(int i = 1; i < N + 1; i += 2){
    int c = 0;
    int j = 1;
    while(i*i <= j){
      if(i*i == j) c++;
      else if(i % j == 0) c += 2;
      j++;
    }
    if(c == 8) count++;
  }
  printf("%d\n", count);
}