output_description: Print a decimal number (or an integer) representing the maximum possible value
  of the last ingredient remaining.
  
  Your output will be judged correct when its absolute or relative error from
  the judge's output is at most 10^{-5}.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      v_1 v_2 \ldots v_N
attempt: A = int(input())
  B = list(map(int, input().split()))
  print(sum(B) / 2 - max(B) / 2)
  
problem_description: Statement
  
  You have a pot and N ingredients. Each ingredient has a real number parameter
  called _value_ , and the value of the i-th ingredient (1 \leq i \leq N) is
  v_i.
  
  When you put two ingredients in the pot, they will vanish and result in the
  formation of a new ingredient. The value of the new ingredient will be (x + y)
  / 2 where x and y are the values of the ingredients consumed, and you can put
  this ingredient again in the pot.
  
  After you compose ingredients in this way N-1 times, you will end up with one
  ingredient. Find the maximum possible value of this ingredient.
samples: [{"input": "2\n    3 4", "output": "3.5\n    \n\nIf you start with two ingredients, the only choice is to put both of them in\nthe pot. The value of the ingredient resulting from the ingredients of values\n3 and 4 is (3 + 4) / 2 = 3.5.\n\nPrinting `3.50001`, `3.49999`, and so on will also be accepted.\n\n* * *"}, {"input": "3\n    500 300 200", "output": "375\n    \n\nYou start with three ingredients this time, and you can choose what to use in\nthe first composition. There are three possible choices:\n\n  * Use the ingredients of values 500 and 300 to produce an ingredient of value (500 + 300) / 2 = 400. The next composition will use this ingredient and the ingredient of value 200, resulting in an ingredient of value (400 + 200) / 2 = 300.\n  * Use the ingredients of values 500 and 200 to produce an ingredient of value (500 + 200) / 2 = 350. The next composition will use this ingredient and the ingredient of value 300, resulting in an ingredient of value (350 + 300) / 2 = 325.\n  * Use the ingredients of values 300 and 200 to produce an ingredient of value (300 + 200) / 2 = 250. The next composition will use this ingredient and the ingredient of value 500, resulting in an ingredient of value (250 + 500) / 2 = 375.\n\nThus, the maximum possible value of the last ingredient remaining is 375.\n\nPrinting `375.0` and so on will also be accepted.\n\n* * *"}, {"input": "5\n    138 138 138 138 138", "output": "138"}]