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<bits/stdc++.h>
//#include <ext/pb_ds/detail/standard_policies.hpp>

const double pi=acos(-1.0);
//memset ( a , 0 , n * sizeof(ll) ) ;
using namespace std;
//using namespace __gnu_pbds;
#define   endl    '\n'
#define   sl(n)     scanf("%lld",&n)
#define   mp      make_pair
#define   pb      push_back
#define   ppb     pop_back
#define   fi      first
#define   se      second
#define   ll      long long
#define   ull     unsigned long long
#define   ld      long double
#define   pii     pair<int, int>
#define   f(i,a,b)  for(ll i = (ll)(a); i < (ll)(b); i++)
#define   rf(i,a,b)   for(ll i = (ll)(a); i > (ll)(b); i--)
#define   ms(a,b)   memset((a),(b),sizeof(a))
#define   abs(x)    ((x<0)?(-(x)):(x))
#define   MAX     1000005
#define   inf     LLONG_MAX
#define   ninf    LLONG_MIN
#define   MIN     INT_MIN
#define   yeet    return 0;
#define fast_io ios_base::sync_with_stdio (false) ; cin.tie(0) ; cout.tie(0) ;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
// Use cout.flush() for interactive problems.
// Use this for increased stack size: g++ -o a.exe -Wl,--stack=256000000 -O2 source.cpp
inline long long  MAX2(long long  a, long long  b){return (a)>(b)?(a):(b);}
inline long long  MAX3(long long  a, long long  b,long long  c){return (a)>(b)?((a)>(c)?(a):(c)):((b)>(c)?(b):(c));}
inline long long  MIN2(long long  a, long long  b){return (a)<(b)?(a):(b);}
inline long long  MIN3(long long  a, long long b,long long c){return (a)<(b)?((a)<(c)?(a):(c)):((b)<(c)?(b):(c));}

//typedef
typedef long int int32;
typedef unsigned long int uint32;
typedef long long int int64;
typedef unsigned long long int  uint64;


int mod = 1e9 +7 ;
int64_t extGcd(int64_t a, int64_t b, int64_t& x, int64_t& y) {if (!a) {x = 0;y = 1;return b;}int64_t x1, y1;int64_t d = extGcd(b % a, a, x1, y1);x = y1 - (b / a) * x1;y = x1;return d;}

inline int addMod(int a, int b, int m = mod) {return ((int64_t)a + b) % m;}
inline int mulMod(int a, int b, int m = mod) {  return ((int64_t)a * b) % m;}
inline int divMod(int a, int b, int m = mod) {int64_t x, y;int64_t g = extGcd(b, m, x, y);if (g != 1) {cerr << "Bad gcd!" << endl;for (;;);}x = (x % m + m) % m;return mulMod(a, x, m);}
int dx[]={1,1,0,-1,-1,-1, 0, 1};
int dy[]={0,1,1, 1, 0,-1,-1,-1};
inline ll exp(ll a,ll b){if(a==0)return 0ll;ll r=1LL;while(b>0){if(b&1){r=r*(a%mod);r=(r+mod)%mod;}b/=2;a=(a%mod)*(a%mod);a=(a+mod)%mod;}return (r+mod)%mod;}
ll gcd(ll a,ll b){if(b==0)return a;if(a==0)return b;return gcd(b,a%b);}
uint32 setbits(ll n){uint32 count=0;while (n){n&=(n-1);count++;}return count; }
//int clearRangeItoJ(int n,int i,int j){ int ones=(~0); int a= ones<<(j+1);int b=(1<<i)-1;int mask=a|b; int ans=n&mask;return ans}

// ll f[MAX];
// ll iv[MAX];
// ll C(ll n,ll r){
//   if(n<r)return 0;
//   ll ans=(f[n]%mod * iv[r]%mod)%mod;
//   ans=(ans%mod * iv[n-r]%mod)%mod;
//   return ans%mod;
// }
// void init(){
//   f[0]=1;
//      iv[0]=1;
//      f(i,1,MAX)f[i]=(i*f[i-1])%mod;
//      iv[MAX-1]=exp(f[MAX-1],mod-2);

//      for(int i=MAX-2;i>=0;--i)iv[i]=(iv[i+1]*(i+1))%mod;

// }

////****************************************************************************************************************************************************************************************************************////
int main(){
     int n;
     cin>>n;
     map<ll,ll>m1;
     for(int i=2;i<=n;i++){
        int num=i;
    
         for(int  j=2;j*j<=num;j++){
                    while(num%j==0){
                         m1[j]++;
                          num=num/j;
                    }                
         }
         if(num>1)m1[ num]++;
     }
     ll cnt3=0;
     ll cnt5=0;
     ll cnt25=0;
     ll cnt15=0;
     ll cnt75=0;
     for(int i=2;i<=n;i++){
         if(m1[i]>=74)cnt75++;
         if(m1[i]>=24)cnt25++;
         if(m1[i]>=14)cnt15++;
         if(m1[i]>=2)cnt3++;
         if(m1[i]>=4)cnt5++;

     }
     ll ans=0;
      ans+= cnt75+(cnt25* (cnt3-1))+(((cnt5*(cnt5-1))/2) *(cnt3-2)  )+(cnt15*(cnt5-1));
     cout<<ans<<endl;


}