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: Wrong Answer)

#include <stdio.h>
#include <math.h>
int main(void){
    int n,m,i,j;
    int a[10],tmp;
    scanf("%d",&n);
    m=n;
    n=pow(2,n);
    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    for(i=0;i<m;i++){
        tmp=pow(2,i);
        for(j=0;j<n;j++){
            if(a[j]>a[j+tmp]){
                a[j]=a[j]-a[j+tmp];
            }else if(a[j]<a[j+tmp]){
                a[j]=a[j+tmp]-a[j];
            }
            a[j+tmp]=0;
            j++;
        }
        n=n/2;
    }
    printf("%d\n",a[0]);
    return 0;
}