output_description: Print the number of trailing zeros in the decimal notation of f(N).
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
attempt: n = float(input())
  a = 0
  if n % 2 == 1:
      print(0)
  else:
      a = n // 596046447753906250
      a += n // 119209289550781250
      a += n // 23841857910156250
      a += n // 4768371582031250
      a += n // 953674316406250
      a += n // 190734863281250
      a += n // 38146972656250
      a += n // 7629394531250
      a += n // 1525878906250
      a += n // 305175781250
      a += n // 305175781250
      a += n // 12207031250
      a += n // 2441406250
      a += n // 488281250
      a += n // 97656250
      a += n // 19531250
      a += n // 3906250
      a += n // 781250
      a += n // 156250
      a += n // 31250
      a += n // 6250
      a += n // 1250
      a += n // 250
      a += n // 50
      a += n // 10
      print(a)
  
problem_description: Statement
  
  For an integer n not less than 0, let us define f(n) as follows:
  
    * f(n) = 1 (if n < 2)
    * f(n) = n f(n-2) (if n \geq 2)
  
  Given is an integer N. Find the number of trailing zeros in the decimal
  notation of f(N).
samples: [{"input": "12", "output": "1\n    \n\nf(12) = 12 \u00d7 10 \u00d7 8 \u00d7 6 \u00d7 4 \u00d7 2 = 46080, which has one trailing zero.\n\n* * *"}, {"input": "5", "output": "0\n    \n\nf(5) = 5 \u00d7 3 \u00d7 1 = 15, which has no trailing zeros.\n\n* * *"}, {"input": "1000000000000000000", "output": "124999999999999995"}]