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)

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

const PRIMES: &'static [u32] = &[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,
    37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97];


fn count_prime(mut n: u32, c: &mut [u32]) {
    for (i, p) in PRIMES.iter().enumerate() {
        let p = *p;
        while n % p == 0 {
            n /= p;
            c[i] += 1;
        }
    }
}

fn main() {
    input!{
        n: u32,
    };

    let mut count = vec![0; PRIMES.len()];
    for i in 2..n + 1 {
        count_prime(i, &mut count);
    }
    // eprintln!("{:?}", count);
    let mut ans = 0;
    for c in count.iter() {
        let c = *c;
        if c >= 74 {
            ans += 1;
        }
    }
    for a in 0..count.len() {
        if count[a] < 2 {
            continue;
        }
        for b in 0..count.len() {
            if count[b] < 4 {
                continue;
            }
            if a != b && count[b] >= 24 {
                ans += 1;
            }
            for c in b + 1..count.len() {
                if count[c] < 4 {
                    continue;
                }
                if a == b || a == c {
                    if count[a] >= 14 {
                        ans += 1;
                    }
                } else {
                    ans += 1;
                }
            }
        }
    }
    
    println!("{}", ans);
}