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: Wrong Answer)

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

namespace AtCorder
{
	public class Program
	{
		static void Main(string[] args)
		{
			int n = int.Parse(Console.ReadLine());

			if (n <= 2) {
				Console.WriteLine(0);
				return;
			}

			var dic = new Dictionary<int, int>();
			for (int i = 2; i <= n; i++) {
				var temp = Prime(i);
				foreach (var key in temp.Keys) {
					if (dic.ContainsKey(key) == false) {
						dic[key] = temp[key];
					} else {
						dic[key] += temp[key];
					}
				}
			}

			int up2 = dic.Where(x => x.Value >= 2).Count();
			int up5 = dic.Where(x => x.Value >= 4).Count();
			if (up2 >= 1 && up5 >= 2) {
				long ans = up2 * (up5 - 1) * (up5 - 2);
				if (up5 >= 3) {
					ans += (up5 - 1) * (up5 - 2) * up5;
				}
				Console.WriteLine(ans);
			} else {
				Console.WriteLine(0);
				return;
			}
		}

		public static Dictionary<int, int> Prime(int n)
		{
			var dic = new Dictionary<int, int>();
			for (int i = 2; i < 10; i++) {
				while (n % i == 0) {
					if (dic.ContainsKey(i) == false) {
						dic[i] = 1;
					} else{
						dic[i]++;
					}

					n /= i;
				}
			}

			return dic;
		}
	}
}