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 JavaScript (node.js v5.12) (judge verdict: Time Limit Exceeded)

function f(k) {
  var j = 1;
  for(var i = 1; i <= k; i++){
    j *= i;
  }
  return j;
}

function get75n(k) {
	var counter = 0
	for (var i = 1; i < N; i++) {
		if (k % i === 0) {
			counter++;
		}
	}
	return counter
}

function main(arg) {
	arg = Number(arg.split("\n")[0])

	var master = [0, 3,5,7]
	var counter = 0;

	N = f(arg)
	for (var i = 1; i < N; i++) {
		if (N % i === 0) {
			if (get75n(i) === 75) {
				counter++;
			}
		}
	}

	console.log(counter)
}
main(require('fs').readFileSync('/dev/stdin', 'utf8'));