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;



namespace CSharpSample01
{



    class Program
    {

        static int[] e;
        
        static int Elen(int m)
        {
            int len = 0;
            for(int i = 0; i < e.Length; i++)
            {
                if (e[i] >= m - 1) len++;
            }

            return len;
        }
        
        static void Main(string[] args)
        {
            int N = int.Parse(Console.ReadLine());

            e = new int[N + 1];

            for(int i = 2; i <= N; i++)
            {
                int now = i;
                for(int j = 2; j <= now; j++)
                {
                    while(now % j == 0)
                    {
                        now /= j;
                        e[j]++;
                    }
                   
                }
            }

            int[] table = { 2, 4, 14, 24, 74 };
            int[] cnt = new int[5];
            
            for(int i = 0; i < e.Length; i++)
            {
                for(int j = table.Length - 1; j >= 0; j--)
                {
                    if(e[i] >= table[j])
                    {
                        cnt[j]++;
                        break;
                    }
                }
            }

            int ans = 0;

            
            ans = Elen(75) + Elen(25) * (Elen(3) - 1) + Elen(15) * (Elen(5) - 1)
                + Elen(5) * (Elen(5) - 1) * (Elen(3) - 2) / 2;

            Console.WriteLine(ans);
        }
    }
}