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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

import math

def fact(n, _factcache={}):
    if n not in _factcache:
        _factcache[n] = math.factorial(n)
    return _factcache[n]

def divisorGenerator(n):
    large_divisors = []
    for i in range(1, int(math.sqrt(n) + 1)):
        if n % i == 0:
            yield i
            if i*i != n:
                large_divisors.append(n / i)
    for divisor in reversed(large_divisors):
        yield divisor


def fiveSeven(n):
    counter = 0
    divisors = list(divisorGenerator(fact(n)))
    for _, divisor in enumerate(divisors):
        if len(list(divisorGenerator(divisor))) == 75:
            counter += 1
    return counter

i = int(input())
print(fiveSeven(i))