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;

typedef long long ll;

template<typename T>
void out(T x) { cout << x << endl; exit(0); }
#define watch(x) cout << (#x) << " is " << (x) << endl

const int maxn = 4e5 + 5;


int n,cnt[105],dp[105];

int main() {
  ios_base::sync_with_stdio(false); cin.tie(0);  cout.tie(0);
  cin>>n;

  for (int i=2; i<=n; i++) {
    int x = i;
    for (int j=2; x>1; j++) {
      while (x%j == 0) {
	cnt[j]++;
	x=x/j;
      }
    }
  }

  dp[1]=1;
  for (int i=1; i<=100; i++) {
    if (cnt[i]>0) {
      for (int j=75; j>=1; j--) {
	for (int x=1; x<=cnt[i]; x++) {
	  if ((x+1)*j <= 75) {
	    dp[(x+1)*j] += dp[j];
	  }
	}
      }
    }
  }

  cout<<dp[75]<<endl;
  return 0;
}