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>
using namespace std;

int main(){
    long long n;
    cin>>n;
    long long a[3*n]; 
    pair<long long,int> p[n*2] , pp[n*2];
    for(int i=0;i<2*n;i++){
        cin>>a[i];
        p[i]=make_pair(a[i],i);
        
    }
    for(int i=2*n;i<3*n;i++){
        cin>>a[i];
    }
    for(int i=0;i<2*n;i++){
        pp[i]=make_pair(a[i+n],i+n);
    }
    sort(pp,pp+2*n);
    int y=1e8;
    long long res=0;
    for(int i=0;i<n;i++){
        res-=pp[i].first;
        y=min(y,pp[i].second);
    }

    vector<long long> vv;
    for(int i=0;i<y;i++){
        vv.push_back(a[i]);
    }
    sort(vv.begin(),vv.end(),greater<long long>());
    for(int i=0;i<n;i++){
        res+=a[i];
    }

    

    sort(p,p+2*n);
    reverse(p,p+2*n);

    int t=0;
    long long ans=0;
    for(int i=0;i<n;i++){
        ans+=p[i].first;
        t=max(t,p[i].second);
    }
    
    vector<long long> v;
    for(int i=t+1;i<3*n;i++){
        v.push_back(a[i]);
    }
    sort(v.begin(),v.end());

    for(int i=0;i<n;i++){
        ans-=v[i];
    }

    cout<<(long long)max(res,ans)<<endl;


}