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)

$N=<>;
for$i(1..$N){
	for$p(grep{(1x$_)!~/^(11+)\1+$/}2..$N){
		$g[$p]+=grep{$i%$p**$_<1}1..6
	}
}
for$a(1..$N){
	$z+=$g[$a]>73;
	for$b(2..$N){
		if($a!=$b){
			$z+=$g[$a]>3&&$g[$b]>13;
			$z+=$g[$a]>1&&$g[$b]>23;
			for$c($b+1..$N){
				if($a!=$c){
					$z+=$g[$a]>1&&$g[$b]>3&&$g[$c]>3;
				}
			}
		}
	}
}
print$z