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++ (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
long long a,b,cnt[100],ans;
void dfs(int x,int a1,bool a2)
{
	if(a1==2&&a2)
	{
		ans++;
		return;
	}
	if(x>97)return;
	dfs(x+1,a1,a2);
	if(cnt[x]>=4&&a1<2)dfs(x+1,a1+1,a2);
	if(cnt[x]>=2&&!a2)dfs(x+1,a1,1);
	if(cnt[x]>=14&&!a2&&a1<2)dfs(x+1,a1+1,1);
	if(cnt[x]>=24&&!a1)dfs(x+1,2,a2);
	if(cnt[x]>=74&&!a1&&!a2)dfs(x+1,2,1);
}
int main()
{
	cin>>a;
	for(int s=2;s<=a;s++)
	{
		b=s;
		for(int v=2;v<=a;v++)
		{
			if(b%v==0)
			{
				cnt[v]++;
				b/=v;
				v--;
			}
		}
	}
	dfs(2,0,0);
	cout<<ans<<endl;
}