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)

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);


	int P[101] = {};
	int N;
	cin >> N;

	rep1(i, N) {
		int tmp = i;
		for (int j = 2; j <= i; j++) {
			while (tmp % j == 0) {
				P[j]++;
				tmp /= j;
			}
		}
	}

	int P2[1001] = {};

	rep1(i, 100) {
		P2[P[i]]++;
	}
	for (int i = 1000; i > 0; i--) {
		P2[i - 1] += P2[i];
	}

	int kotae = P2[74];
	kotae += (P2[2] - 1) * P2[24];
	kotae += (P2[4] - 1) * P2[14];
	kotae += (P2[2] - 2) * (P2[4] - 1) * P2[4] / 2;

	co(kotae);

	Would you please return 0;
}