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.

Examples

Input

2
1
3
10
19

Output

7

Input

3
1
3
2
4
6
8
100
104

Output

2

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