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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define REP(i,n) for(int i=0;i<n;i++)
#define LOOP(i,x,n) for(int i=x;i<n;i++)
#define ALL(v) (v).begin(),(v).end()
#define PB push_back
#define MP make_pair
#define FIR first
#define SEC second
#define int long long
using namespace std;
const int MOD=1000000007;
const int INF=1000000009;


signed main(){
  int n;
  cin>>n;
  vector<int> v;
  REP(i,pow(2,n)){
    int a;
    cin>>a;
    v.PB(a);
  }
  REP(i,n){
    vector<int> k;
    REP(j,pow(2,n-i)){
      int tmp;
      if(v[j]==v[j+1]){
        tmp=v[j];
      }else{
       tmp=abs(v[j]-v[j+1]);
      }
      k.PB(tmp);
      j++;
    }
    REP(j,k.size()){
      v[j]=k[j];
    }
  }
  cout<<v[0]<<endl;
  return 0;
}