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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

from math import sqrt

def cnt_yakusu(num):
    sq_num = int(sqrt(num))
    cnt = 0
    for i in range(1, sq_num):
        for j in range(sq_num+1, num+1):
            if i*j==N:
                cnt += 2
    for i in range(350):
        if i**2 == num:
            cnt += 1
    return num

N = int(input())
ans = 0
for n in range(N, 1, -1):
    cnt = cnt_yakusu(n)
    if n == 75:
        ans += 1
print(ans)