output_description: Print the eventual hardness of the last stone remaining.
  
  * * *
input_description: The input is given from Standard Input in the following format:
  
      
      
      N
      A_1
      A_2
      :
      A_{2^N}
attempt: 3
  1
  3
  2
  4
  6
  8
  100
  104
  
problem_description: Statement
  
   _Kode Festival_ is an anual contest where the hardest stone in the world is
  determined. (Kode is a Japanese word for "hardness".)
  
  This year, 2^N stones participated. The hardness of the i-th stone is A_i.
  
  In the contest, stones are thrown at each other in a knockout tournament.
  
  When two stones with hardness X and Y are thrown at each other, the following
  will happen:
  
    * When X > Y: The stone with hardness Y will be destroyed and eliminated. The hardness of the stone with hardness X will become X-Y.
  
    * When X = Y: One of the stones will be destroyed and eliminated. The hardness of the other stone will remain the same.
  
    * When X < Y: The stone with hardness X will be destroyed and eliminated. The hardness of the stone with hardness Y will become Y-X.
  
  The 2^N stones will fight in a knockout tournament as follows:
  
    1. The following pairs will fight: (the 1-st stone versus the 2-nd stone), (the 3-rd stone versus the 4-th stone), ...
  
    2. The following pairs will fight: (the winner of (1-st versus 2-nd) versus the winner of (3-rd versus 4-th)), (the winner of (5-th versus 6-th) versus the winner of (7-th versus 8-th)), ...
  
    3. And so forth, until there is only one stone remaining.
  
  Determine the eventual hardness of the last stone remaining.
samples: [{"input": "2\n    1\n    3\n    10\n    19", "output": "7\n    \n\n* * *"}, {"input": "3\n    1\n    3\n    2\n    4\n    6\n    8\n    100\n    104", "output": "2"}]