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>
int main(void){
  int N,st[300000],i,j,k,l,m,n;
  scanf("%d",&N);
  j=1;
  k=1;
  m=2;
  for(i=1;i<=N;i++){
    j=j*2;
  }
  for(i=0;i<j;i++){
    scanf("%d",&st[i]);
  }
  for(l=1;l<=N;l++){
    for(i=0;i<j;i=i+m){
      if(st[i]<st[i+k]){
	n=st[i];
	st[i]=st[i+k];
	st[i+k]=n;
      }
      if(st[i]!=st[i+k]){
	st[i]=st[i]-st[i+k];
      }
    }
  k=k*2;
  m=m*2;
}
printf("%d\n",st[0]);
return 0;
}