output_description: Print the maximum number of times the operation can be applied.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
attempt: # 入力
  N = input()
  N = int(N)
  
  # 素因数分解
  L = []
  M = []
  Nu = N
  for i in range(2, int(-(-(N**0.5) // 1)) + 1):
      if Nu % i == 0:
          j = 0
          while Nu % i == 0:
              j += 1
              Nu //= i
          # L.append(i)
          M.append(j)
  
  if Nu != 1:
      # L.appnend(Nu)
      M.append(1)
  
  if M == []:
      if N != 1:
          # L.append(N)
          M.append(1)
      else:
          M.append(0)
  
  # 階差数列
  l = max(M)
  p = 0
  P = []
  for i in range(1, l + 1):
      p += i
      P.append(p)
  
  # 計算
  Answer = 0
  l = len(M)
  for i in range(l):
      m = M[i]
      if m != 0:
          while not (m in P):
              m -= 1
          Index = P.index(m)
          Index += 1
          Answer += Index
  
  print(Answer)
  
problem_description: Statement
  
  Given is a positive integer N. Consider repeatedly applying the operation
  below on N:
  
    * First, choose a positive integer z satisfying all of the conditions below:
      * z can be represented as z=p^e, where p is a prime number and e is a positive integer;
      * z divides N;
      * z is different from all integers chosen in previous operations.
    * Then, replace N with N/z.
  
  Find the maximum number of times the operation can be applied.
samples: [{"input": "24", "output": "3\n    \n\nWe can apply the operation three times by, for example, making the following\nchoices:\n\n  * Choose z=2 (=2^1). (Now we have N=12.)\n  * Choose z=3 (=3^1). (Now we have N=4.)\n  * Choose z=4 (=2^2). (Now we have N=1.)\n\n* * *"}, {"input": "1", "output": "0\n    \n\nWe cannot apply the operation at all.\n\n* * *"}, {"input": "64", "output": "3\n    \n\nWe can apply the operation three times by, for example, making the following\nchoices:\n\n  * Choose z=2 (=2^1). (Now we have N=32.)\n  * Choose z=4 (=2^2). (Now we have N=8.)\n  * Choose z=8 (=2^3). (Now we have N=1.)\n\n* * *"}, {"input": "1000000007", "output": "1\n    \n\nWe can apply the operation once by, for example, making the following choice:\n\n  * z=1000000007 (=1000000007^1). (Now we have N=1.)\n\n* * *"}, {"input": "997764507000", "output": "7"}]