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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <memory>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <sstream>
#include <unordered_map>
#include <vector>


#define INF 4000000000000000000LL
#define MOD 1000000007
#define ALL(x) std::begin(x), std::end(x)


int main(int argc, char** argv)
{
  std::cin.tie(0);
  std::ios_base::sync_with_stdio(0);

  std::cout << std::fixed << std::setprecision(6);
  std::cerr << std::fixed << std::setprecision(6);

  int N, C = 0;

  std::cin >> N;

  for (int i = 1; i <= N; i += 2) {
    int c = 0;
    
    for (int j = 1; j <= i; j ++)
      if (i % j == 0)
        c ++;

    if (c == 8)
      C ++;
  }

  std::cout << C << std::endl;
  
  return 0;
}