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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
unsigned prime[15] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47}, over[5], divisor[5] = {3,5,15,25,75};

int main(){
    unsigned N;
    scanf("%u", &N);
    unsigned power, _N;
    for (size_t i = 0; i < 15; i++){
        power = 1, _N = N;
        while(_N /= prime[i]) power += _N;
        for (size_t j = 0; j < 5; j++){
            if(power >= divisor[j]) over[j]++;
        }
    }
    printf("%u\n", over[4] + over[3]*(over[0] - 1) + over[2]*(over[1] - 1) + (over[1]*(over[1] - 1)>>1)*(over[0] - 2));
	return 0;
}