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: Wrong Answer)

// template version 1.6
// {{{ define basic macro
using namespace std;
#include <iostream>
#include <bits/stdc++.h>
// varibable settings
#define int long long
#define INF 1000000000000000000LL
// #define INF 2147483647
#define MOD 1000000007LL
// #define MOD 998244353LL
#define infile "../test/sample-1.in"

#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define _rrep(i,n) rrepi(i,0,n)
#define rrepi(i,a,b) for(int i=(int)(b-1);i>=(int)(a);--i)
#define rrep(...) _overload3(__VA_ARGS__,rrepi,_rrep,)(__VA_ARGS__)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define uni(x) sort(all(x));x.erase(unique(all(x)),x.end())
#define ten(n) ((int)1e##n)
template<class T, size_t N> size_t sza(const T (&array)[N]) { return N; }
template<class T> inline void chmax(T &a, const T &b) { if(a < b) a = b; }
template<class T> inline void chmin(T &a, const T &b) { if(a > b) a = b; }
template<class T=int> T in() {T x; cin>>x; return (x);}
struct Fast { Fast(){ std::cin.tie(0); ios::sync_with_stdio(false); } } fast;

// dump macro
#ifdef PCM
  #include "dump.hpp"
#else
  #define dump(...) 42
  #define dump_1d(...) 42
  #define dump_2d(...) 42
#endif

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef long double ld;
typedef pair<int,int> pii;
typedef tuple<int,int,int> iii;
template<typename T> using PQ = priority_queue<T, vector<T>, greater<T>>;
int dx[]={1, -1, 0, 0};
int dy[]={0, 0, 1, -1};
#define fi first
#define se second

// }}}

map<int, int> pf;

void prime_factor(int n) { // 素因数分解 (o(√N))
    for(int i = 2; i * i <= n; i++) {
        while(n % i == 0) {
            ++pf[i];
            n /= i;
        }
    }
    if (n != 1) pf[n] += 1;
}

int solve(){
  int N;cin>>N;
  rep(i, 2, N+1){
    prime_factor(i);
  }
  dump(pf);

  map<int, int> c;
  for (auto el:pf){
    if (el.second>=75) c[75]++;
    if (el.second>=24) c[24]++;
    if (el.second>=14) c[14]++;
    if (el.second>=4) c[4]++;
    if (el.second>=2) c[2]++;
  }
  int res=0;
  res += c[75];
  res += c[24]*(c[2]-1);
  res += c[14]*(c[4]-1);
  if (c[4]>=2 and c[2]>=3) res+=c[4]*(c[4]-1)/2*(c[2]-2);
  cout << res << endl;

  return 0;
}

signed main() { //{{{
#ifdef INPUT_FROM_FILE
  std::ifstream in(infile);
  std::cin.rdbuf(in.rdbuf());
#endif
  solve();
  return 0;
} //}}}