Score : $100$ points

Problem 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.

Constraints

• $1 \leq N \leq 18$
• $1 \leq A_i \leq 10^9$
• $A_i$ is an integer.

Input

The input is given from Standard Input in the following format:

$N$

$A_1$

$A_2$

:

$A_{2^N}$

Output

Print the eventual hardness of the last stone remaining.

2
1
3
10
19

7

3
1
3
2
4
6
8
100
104

2