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;

typedef long long ll;
using ll = long long;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

    int n;cin>>n;
    priority_queue<ll,vector<ll>,greater<ll>> q;
    ll su=0;
    vector<ll> a(3*n),ma(3*n),mi(3*n);
    for(int i=0;i<3*n;i++) cin>>a[i];
    for(int i=0;i<n;i++){
        q.push(a[i]);
        su+=a[i];
        ma[i]=su;
    }
    for(int i=n;i<2*n;i++){
        q.push(a[i]);
        su+=a[i];
        su-=q.top();q.pop();
        ma[i]=su;
    }
    priority_queue<ll> q2;
    su=0;
    for(int i=3*n-1;i>=2*n;i--){
        q2.push(a[i]);
        su+=a[i];
        mi[i]=su;
    }
    for(int i=2*n-1;i>=n;i--){
        q2.push(a[i]);
        su+=a[i];
        su-=q2.top();q2.pop();
        mi[i]=su;
    }
    ll res=-10000000000;
    for(int i=n-1;i<2*n;i++){
        res=max(res,ma[i]-mi[i+1]);
        //cout<<i<<" "<<ma[i]<<" "<<mi[i+1]<<endl;
    }
    cout<<res<<endl;
}