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 <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <list>
#include <cmath>

using namespace std;
typedef long long ll;
using Pll = pair<ll,ll>;
const ll inf = 1LL<<60;

int main() {
    ll N;cin >> N;
    priority_queue<ll> L;
    priority_queue<ll,vector<ll>,greater<ll>>F;
    vector<ll> M(N);
    ll fval=0,lval=0;
    for(int i=0;i<N;i++){
        ll a;
        cin >> a;
        F.push(a);
        fval+=a;
    }
    for(int i=0;i<N;i++){
        cin >> M[i];
        F.push(M[i]);
    }
    for(int i=0;i<N;i++){
        ll a;
        cin >> a;
        L.push(a);
        lval+=a;
    }
    vector<ll> resf(N+1),resl(N+1);
    resf[0]=fval;
    resl[N]=lval;
    cout << fval << " " << lval << endl;
    for(int i=0;i<N;i++){
            fval+=M[i];
            F.push(M[i]);
            ll tf=F.top();
            fval-=tf;
            F.pop();
            resf[i+1]=fval;

            lval+=M[N-1-i];
            L.push(M[N-1-i]);
            ll tl = L.top();
            lval-=tl;
            L.pop();
            resl[N-i-1]=lval;
    }
    ll ans=-inf;
    for(int i=0;i<=N;i++){
        
        ans=max(ans,resf[i]-resl[i]);
        //cout << resf[i] << " " << resl[i] << endl;
    }
    
    cout << ans << endl;
    return 0;
}