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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;

class D
{
	static void Main()
	{
		var n = int.Parse(Console.ReadLine());

		var ps = new List<int>();
		for (int i = 2; i <= n; i++) if (ps.All(p => i % p != 0)) ps.Add(i);

		var powers = ps.ToDictionary(p => p, p => 1);
		foreach (var p in ps)
			for (int i = 1; i <= n; i++)
			{
				var t = i;
				while (t % p == 0)
				{
					t /= p;
					powers[p]++;
				}
			}

		Func<int, int> countGeq = v => powers.Count(x => x.Value >= v);
		var r = countGeq(75) +
			countGeq(25) * (countGeq(3) - 1) +
			countGeq(15) * (countGeq(5) - 1) +
			countGeq(5) * (countGeq(5) - 1) / 2 * (countGeq(3) - 2);
		Console.WriteLine(r);
	}
}