Problem:
Score : 400 points

Problem Statement

You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?

Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.

Note

When a positive integer A divides a positive integer B, A is said to a divisor of B.
For example, 6 has four divisors: 1, 2, 3 and 6.

Constraints

1 \leq N \leq 100

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of the Shichi-Go numbers that are divisors of N!.

Sample Input 1

9

Sample Output 1

0

There are no Shichi-Go numbers among the divisors of 9! = 1 \times 2 \times ... \times 9 = 362880.

Sample Input 2

10

Sample Output 2

1

There is one Shichi-Go number among the divisors of 10! = 3628800: 32400.

Sample Input 3

100

Sample Output 3

543

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void){
  int N;
  int N_fact = 1;//Nの階乗
  int i, j, k;//カウンタ
  int divisor;//約数の個数
  int shitigosuu = 0;//七五数の個数

  scanf("%d", &N);

  //Nの階乗を求める
  for(i = 1; i <= N; i++){
    N_fact *= i;
  }

  for(i = 1; i <= N_fact; i++){
    divisor = 0;
    for(j = 1; j <= i; j++){
      if(i%j == 0){
        divisor++;
      }
    }
    if(divisor == 75){
      shitigosuu++;
    }
  }

  printf("%d\n", shitigosuu);
  return 0;
}