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 <iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
#include<set>
#include<queue>
#include<map>
#include<cmath>
using namespace std;
#define LL long long
#define rep(i,a,n) for(int i=a;i<n;++i)
#define per(i,a,n) for(int i=n-1;i>=a;--i)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define all(v) v.begin(),v.end()
#define sz(v) ((int)sizeof(v))
#define N 1000005
int fun(int a,int b){
    if(a!=b) return abs(a-b);
    return a;
}
int main()
{
    int i,n,t;
    queue<int>q;
    scanf("%d",&n);
    for(i=0;i<(1<<n);++i){
        scanf("%d",&t);
        q.push(t);
    }
    int a,b;
    while(q.size()>1){
        a=q.front();
        q.pop();
        b=q.front();
        q.pop();
        int t=fun(a,b);
        q.push(t);
    }
    cout<<q.front()<<endl;
    return 0;
}