
Score : 100 points
Problem StatementKode 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:
The following pairs will fight: (the 1-st stone versus the 2-nd stone), (the 3-rd stone versus the 4-th stone), ...
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)), ...
And so forth, until there is only one stone remaining.
Determine the eventual hardness of the last stone remaining.
Constraints
1 \leq N \leq 18
1 \leq A_i \leq 10^9
A_i is an integer.
InputThe input is given from Standard Input in the following format:
N
A_1
A_2
:
A_{2^N}
OutputPrint the eventual hardness of the last stone remaining.
Sample Input 12
1
3
10
19
Sample Output 17
Sample Input 23
1
3
2
4
6
8
100
104
Sample Output 22
