Problem:
Score : 500 points

Problem Statement

Let N be a positive integer.

There is a numerical sequence of length 3N, a = (a_1, a_2, ..., a_{3N}).
Snuke is constructing a new sequence of length 2N, a', by removing exactly N elements from a without changing the order of the remaining elements.
Here, the score of a' is defined as follows: (the sum of the elements in the first half of a') - (the sum of the elements in the second half of a').

Find the maximum possible score of a'.

Constraints

1 ≤ N ≤ 10^5

a_i is an integer.

1 ≤ a_i ≤ 10^9

Partial Score

In the test set worth 300 points, N ≤ 1000.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_{3N}

Output

Print the maximum possible score of a'.

Sample Input 1

2
3 1 4 1 5 9

Sample Output 1

1

When a_2 and a_6 are removed, a' will be (3, 4, 1, 5), which has a score of (3 + 4) - (1 + 5) = 1.

Sample Input 2

1
1 2 3

Sample Output 2

-1

For example, when a_1 are removed, a' will be (2, 3), which has a score of 2 - 3 = -1.

Sample Input 3

3
8 2 2 7 4 6 5 3 8

Sample Output 3

5

For example, when a_2, a_3 and a_9 are removed, a' will be (8, 7, 4, 6, 5, 3), which has a score of (8 + 7 + 4) - (6 + 5 + 3) = 5.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define ll          long long
#define all(a)      (a).begin(),(a).end()
using namespace std;
//ll i,j,k,l,n,x;
//string s,s1;
int main()
{
    int n ;
    cin>>n ;
    vector<ll>v ;
    for(int i =0 ;i<3*n ; i++)
    {
        ll x ;
        cin>>x ;
        v.push_back(x) ;
    }
    ll al[n+1] ;
    ll ar[n+1] ;
    priority_queue<ll,vector<ll>,greater<ll>>p ;
    priority_queue<ll>q ;
    ll sum = 0 ;
    ll j =0 ;
    while(j<n)
    {
        p.push(v[j]) ;
        sum = sum+v[j] ;
        j++ ;

    }
    al[0] = sum ;
    ll i = 1 ;
    while(j<2*n)
    {
        if(v[j]>p.top())
        {
            sum = sum - p.top() + v[j] ;
            p.pop() ;
            p.push(v[j]) ;

        }
        al[i] = sum ;
        i++ ;
        j++ ;
    }
    int k = v.size()-1 ;
    //int f = n ;
    sum = 0 ;
    while(k>((2*n)-1))
    {   q.push(v[k]) ;
        sum = sum+v[k] ;
        k-- ;
    }
    int f = n-1 ;
    al[n]=sum ;
    while(k>(n-1))
    {
        if(v[k]<q.top())
        {
            sum = sum - q.top() + v[k] ;
            q.pop() ;
            q.push(v[k]) ;
        }
        ar[f] = sum ;
        f-- ;
        k-- ;
    }
    k =0 ;
    ll ans = al[k]-ar[k] ;
    k++  ;
    while(k<=n)
    {
        ans = max(ans,al[k]-ar[k]) ;
        k++ ;
    }
    cout<<ans ;
}