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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.lang.*;


public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int so[] = new int[n + 1];
        int count = 0;
        for (int i = 2; i <= n; i++) {
            int k = i;
            for (int j = 2; j <= i; j++) {
                while (k % j == 0) {
                    so[j]++;
                    k /= j;
                }
            }
        }
        Arrays.sort(so);
        int a = 0;
        int b = 0;
        int c = 0;
        int d = 0;
        int e = 0;
        for (int i = 1; i <= n + 1; i++) {
            if (so[i - 1] < 2) {
                a = i;
            }
            if (so[i - 1] < 4) {
                b = i;
            }
            if (so[i - 1] < 14) {
                c = i;
            }
            if (so[i - 1] < 24) {
                d = i;
            }
            if (so[i - 1] < 74) {
                e = i;
            }
        }
        count += n + 1 - e;
        count += (n + 1 - d) * (n + 1 - a - 1);
        count += (n + 1 - c) * (n + 1 - b - 1);
        count += (n + 1 - b) * (n + 1 - b - 1) * (n + 1 - a - 2) / 2;
        System.out.println(count);
    }
}