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())
M = [0] * (101)

def prime_factorize(n):
    a = []
    while n % 2 == 0:
        a.append(2)
        n //= 2
    f = 3
    while f * f <= n:
        if n % f == 0:
            a.append(f)
            n //= f
        else:
            f += 2
    if n != 1:
        a.append(n)
    return a

for i in range(2, N + 1):
  for m in prime_factorize(i):
    M[m] += 1
#print(M[:N+1])

n3, n5, n15, n25, n75 = 0, 0, 0, 0, 0
for i in range(N+1):
  if M[i] >= 2:  n3  += 1
  if M[i] >= 4:  n5  += 1
  if M[i] >= 14: n15 += 1
  if M[i] >= 24: n25 += 1
  if M[i] >= 74: n75 += 1
#print(n3, n5, n15, n25, n75)   

answer = n75 + n25 * (n3 - 1) + n15 * (n5 - 1) + n5 * (n5 - 1) * (n3 - 2) // 2
print(answer)