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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>

int f(int X, int Y){
	if(X > Y){
		return X - Y;
	}
	else if(X < Y){
		return Y - X;
	}
	else{
		return X;
	}
}

int main(){
	int N, i, j;
	scanf("%d", &N);
	int **A = (int **)malloc(sizeof(int) * (N + 1));
	for(i = 0; i <= N; i++){
		A[i] = (int *)malloc(sizeof(int) * (1 << i));
	}
	for(j = 0; j < (1 << N); j++){
		scanf("%d", &A[N][j]);
	}
	for(i = N - 1; i >= 0; i--){
		for(j = 0; j < (1 << i); j++){
			A[i][j] = f(A[i + 1][2 * j], A[i + 1][2 * j + 1]);
		}
	}
	printf("%d\n", A[0][0]);
	return 0;
}