p03213 AtCoder Beginner Contest 114 - 756_16115
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.

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!.

Examples

Input

9


Output

0


Input

10


Output

1


Input

100


Output

543
#include <bits/stdc++.h>
#define rp(i,n) for(int i=0; i<n; i++)
using namespace std;
using ll=long long;
using P=pair<int,int>;
using G=vector<vector<int>>;

int howmany(int x,int p){
    if(x<p) return 0;
    return x/p+howmany(x/p,p);
}
int main(){
    int n;
    cin >> n;
    int prime[15]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47};
    int a=0,b=0,c=0,d=0,e=0;
    for(int i=0; i<15; i++){
        int m=howmany(n,prime[i]);
        if(m>=2) a++;
        if(m>=4) b++;
        if(m>=14) c++;
        if(m>=24) d++;
        if(m>=74) e++;
    }
    int ans=(a-2)*b*(b-1)/2+(b-1)*c+(a-1)*d+e;
    cout << ans << endl;
    return 0;
}