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 (2.4.0) (judge verdict: Accepted)

N = int(input())
from math import ceil, sqrt, factorial
from itertools import permutations
def eratosthenes(n):
    l = [1 for i in range(n+1)]
    l[0] = l[1] = 0
    for i in range(2, ceil(sqrt(n))):
        if l[i] == 0: continue
        for j in range(i*2, n+1, i):
            l[j] = 0
    return l

m = factorial(N)
ans = set()

p = eratosthenes(530)

p = [i for i,v in enumerate(p) if v ]

for i in p:
    a = i**74
    if not m % a:
        ans.add(a)
        break

for a,b in permutations(p,2):
    c = a**2*b**24
    if not m % c:
        ans.add(c)
    c = a**4*b**14
    if not m % c:
        ans.add(c)

for a,b,c in permutations(p, 3):
    d = a**2 * b**4 * c ** 4
    if not m % d:
        ans.add(d)

print(len(ans))