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: Runtime Error)

N = int(input())

fact_ct = [0 for _ in range(N+1)]
pr = []

def primes():
  ans = set()
  visited = set()
  for p in range(2,N+1):
    if p in visited:
      continue
    ans.add(p)
    q = p
    while q <= N:
      visited.add(q)
      q += p
    
def add_fact_ct(n):
  visited = set()
  for p in range(2,N+1):
    if p not in pr:
      continue
    while n % p == 0:
      fact_ct[p] += 1
      n //= p 
    n //= p
  
pr = primes() 
for n in range(2, N+1):
  add_fact_ct(n)
  
ct_75 = 0
ct_25 = 0
ct_15 = 0
ct_5 = 0
ct_3 = 0
ct_1 = 0
  
for ct in fact_ct:
  if ct >= 75:
    ct_75 += 1
  if ct >= 25:
    ct_25 += 1
  if ct >= 15:
    ct_15 += 1
  if ct >= 5:
    ct_5 += 1 
  if ct >= 3:
    ct_3 += 1
  if ct >= 1:
    ct_1 += 1

ans = 0

ans += ct_75 * (ct_1-1)
ans += ct_25 * (ct_3-1)
ans += ct_15 * (ct_5-1)

print(ans)