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 <iostream>
#include <vector>
using namespace std;

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

    int N; cin >> N;
    vector<int> v(N+1), c(5);
    for (int i = 2; i <= N; i++) {
        int x = i;
        for (int j = 2; j <= i; j++) {
            while (x % j == 0) {
                v[j]++;
                if (v[j] == 74) c[0]++;
                else if (v[j] == 24) c[1]++;
                else if (v[j] == 14) c[2]++;
                else if (v[j] == 4) c[3]++;
                else if (v[j] == 2) c[4]++;
                x /= j;
            }
        }
    }

    int ans = 0;
    ans += c[0];
    ans += c[1] * (c[4]-1);
    ans += c[2] * (c[3]-1);
    ans += c[3] * (c[3]-1) / 2 * (c[4]-2);
    cout << ans << "\n";
}