Problem:
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:

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.

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.

Sample Input 1

2
1
3
10
19

Sample Output 1

7

Sample Input 2

3
1
3
2
4
6
8
100
104

Sample Output 2

2

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;
public class Main {
	public static void main(String[] args){
		Scanner as = new Scanner(System.in);
		int n = Integer.parseInt(as.next());
		int l = n;
		int[] a;
		int b = 0;
		int c = (int)Math.pow(2, n);
		a = new int[c];
		for(int i=0;i<c;i++){
			a[i]  = Integer.parseInt(as.next());
		}
		for(int s=0;s<n;s++){
			for(int i=0;i<(int)(Math.pow(2, l));i+=2){
				if(a[i]<a[i+1]){
					a[i+1]-=a[i];
					a[b]=a[i+1];
					b+=1;
				}else if(a[i]>a[i+1]){
					a[i]-=a[i+1];
					a[b]=a[i];
					b+=1;
				}else{
					a[b]=a[i];
					b+=1;
				}
			}
			l-=1;
			b = 0;
		}
		System.out.println(a[0]);
	}
}