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,N = 1;
  int i,j,k,l;
  int count = 0,result = 0;
  
  scanf("%d",&n);
  
  for(i = 1;i < n;i++){
    N *= i;
  }
  
  for(j = 1;j <= N;j++){
    k = N % j;
    if(k == 0){
      count = 0;
      for(l = 1;l <= k;l++){
        if(k % l == 0){
         count++; 
        }
      }
      if(count == 75){
       result++; 
      }
    }
  }
      
  printf("%d",result);
  return 0;
}