output_description: Print the value of R_3.
  
  The output is considered correct if the absolute or relative error is at most
  10^{-6}.
  
  * * *
input_description: The input is given from Standard Input in the following format:
  
      
      
      R_1 R_2
attempt: import math, string, itertools, fractions, heapq, collections, re, array, bisect, sys, random, time, copy, functools
  
  sys.setrecursionlimit(10**7)
  inf = 10**20
  eps = 1.0 / 10**15
  mod = 10**9 + 7
  
  
  def LI():
      return [int(x) for x in sys.stdin.readline().split()]
  
  
  def LI_():
      return [int(x) - 1 for x in sys.stdin.readline().split()]
  
  
  def LF():
      return [float(x) for x in sys.stdin.readline().split()]
  
  
  def LS():
      return sys.stdin.readline().split()
  
  
  def I():
      return int(sys.stdin.readline())
  
  
  def F():
      return float(sys.stdin.readline())
  
  
  def S():
      return input()
  
  
  def pf(s):
      return print(s, flush=True)
  
  
  def main():
      a, b = LI()
  
      return 1 / (1 / a + 1 / b)
  
  
  print(main())
  
problem_description: Statement
  
  In an electric circuit, when two resistors R_1 and R_2 are connected in
  parallel, the equivalent resistance R_3 can be derived from the following
  formula:
  
    * \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
  
  Given R_1 and R_2, find R_3.
samples: [{"input": "2 3", "output": "1.2000000000\n    \n\n* * *"}, {"input": "100 99", "output": "49.7487437186"}]