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
import java.util.*;

public class Main {
    static Scanner scanner = new Scanner(System.in);

    static long[] primeFactors;
    public static void main(String[]$) {
        int n = scanner.nextInt();
        primeFactors = new long[n];
        for (int i = 2; i <= n; i++) {
            for (int j = 2, c = i; j <= i && c != 1; j++) {
                while (c % j == 0) {
                    primeFactors[j - 1]++;
                    c /= j;
                }
            }
        }
        System.out.println(count(75) + count(25) * (count(3) - 1) + count(15) * (count(5) - 1) + count(5) * (count(5) - 1) * (count(3) - 2) / 2);
    }

    static long count(int n) {
        return Arrays.stream(primeFactors).filter(i -> i >= n - 1).count();
    }
}