output_description: Print the number of the Shichi-Go numbers that are divisors of N!.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
attempt: import sys
  input = lambda: sys.stdin.readline().rstrip()
  
  N = int(input())
  
  e = [0] * (N+1)
  
  for i in range(2, N+1):
      cur = i
      for j in range(2, i+1):
          while cur % j == 0:
              e[j] += 1
              cur //= j
  
  def num(n):
      return len(list(filter(lambda x: x >= n-1, e)))
  
  # num(3)-1はnum(25)で3以上のものを一個使っているので、一つ引いている。//2は、例えばa^4 * b^4 * c^2 と b^4 * a^4 * c^2がかぶって計上されるため。
  print(num(75) + num(25)*(num(3)-1) + num(15)*(num(5)-1) + num(5)*(num(5)-1)*(num(3)-2)//2)
problem_description: 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.
samples: [{"input": "9", "output": "0\n    \n\nThere are no Shichi-Go numbers among the divisors of 9! = 1 \\times 2 \\times\n... \\times 9 = 362880.\n\n* * *"}, {"input": "10", "output": "1\n    \n\nThere is one Shichi-Go number among the divisors of 10! = 3628800: 32400.\n\n* * *"}, {"input": "100", "output": "543"}]