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 Rust (1.15.1) (judge verdict: Accepted)

// Practice AtCoder
//author: Leonardone @ NEETSDKASU

// 解説読後

fn main() {
	let mut stdin = String::new();
    std::io::Read::read_to_string(
    	&mut std::io::stdin(),
        &mut stdin).unwrap();
    let mut stdin = stdin.split_whitespace();
    let mut get = || stdin.next().unwrap();
    macro_rules! get {
    	($t:ty) => (get().parse::<$t>().unwrap());
        () => (get!(i64));
    }
    
    let n =  get!(usize);
    
    let mut f = vec![0; n+1];
    for j in 2..n+1 {
    	let mut t = j;
        for i in 2..j+1 {
	    	while t % i == 0 {
    	    	f[i] += 1;
        	    t /= i;
        	}
		}
    }
    let mut ans = 0;
    for i in 0..f.len() {
    	if f[i] >= 74 {
        	ans += 1;
        }
    	for j in 0..f.len() {
        	if i == j {
            	continue;
            }
            if f[i]>=14&&f[j]>=4 {
            	ans += 1;
            }
            if f[i]>=24&&f[j]>=2 {
            	ans += 1;
            }
        	for k in j+1..f.len() {
            	if i == k {
                	continue;
                }
            	if f[i]>=2&&f[j]>=4&&f[k]>=4 {
                	ans += 1;
                }
            }
        }
    }
    // println!("{:?}", f);
    println!("{}", ans);
}