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 Perl (v5.18.2) (judge verdict: Accepted)

$N=<>;
@P=grep{(1x$_)!~/^(11+)\1+$/}2..$N;
for$i(1..$N){
	for$p(@P){
		$c=0;
		while($i%$p==0){
			$i/=$p;
			++$c
		}
		$c[$p]+=$c
	}
}
$z=0;
for$a(2..$N){
	$z+=$c[$a]>=74;
	for$b(2..$N){
		if($a!=$b){
			$z+=$c[$a]>=4&&$c[$b]>=14;
			$z+=$c[$a]>=2&&$c[$b]>=24;
			for$c($b+1..$N){
				if($a!=$c){
					$z+=$c[$a]>=2&&$c[$b]>=4&&$c[$c]>=4;
				}
			}
		}
	}
}
print$z