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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>

#include <iostream>
#include <complex>
#include <string>
#include <algorithm>
#include <numeric>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>

#include <functional>
#include <cassert>

typedef long long ll;
using namespace std;

#define debug(x) cerr << __LINE__ << " : " << #x << " = " << (x) << endl;

#define mod 1000000007 //1e9+7(prime number)
#define INF 1000000000 //1e9
#define LLINF 2000000000000000000LL //2e18
#define SIZE 200010



int main(){
  int counter[110] = {};
  int n;

  cin >> n;
  
  for(int i=1;i<=n;i++){
    int t = i;
    for(int j=2;j*j<=t;j++){
      while(t%j == 0){
        counter[j]++;
        t /= j;
      }
    }
    counter[t]++;
  }
  
  int ans = 0;
  
  for(int i=2;i<=n;i++){
    ans += counter[i] >= 74;
    for(int j=2;j<=n;j++){
      // 25 * 3, 15 * 3
      if(i == j) continue;
      ans += counter[i] >= 4 && counter[j] >= 14;
      ans += counter[i] >= 2 && counter[j] >= 24;
      for(int k=i+1;k<=n;k++){
        // 5 * 5 * 3
        if(i == k || j == k) continue;
        ans += counter[i] >= 4 && counter[k] >= 4 && counter[j] >= 2;
      }
    }
  }

  cout << ans << endl;
  
  return 0;
}