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.Linq;
using System.Collections.Generic;

namespace Algorithm
{
    class Program
    {
        static int[] e;
        static void Main(string[] args)
        {
            var n = int.Parse(Console.ReadLine());
            e = new int[n + 1];

            foreach (var i in Enumerable.Range(2, n  - 1))
            {
                var cur = i;
                foreach ( var j in Enumerable.Range(2, i - 1))
                {
                    while (cur % j == 0)
                    {
                        e[j]++;
                        cur /= j;
                    }
                }
            }

            var ans = Num(75) + Num(25) * (Num(3) - 1) + Num(15) * (Num(5) - 1) + Num(5) * (Num(5) - 1) * (Num(3) - 2) / 2;
            Console.WriteLine(ans);
        }

        static int Num(int m)
        {
            return e.Count(z => z >= m - 1);
        }
    }
}