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: Accepted)

#include <bits/stdc++.h>
#define rep(i,n,m) for(int i=n;i<(int)(m);i++)
#define rrep(i,n,m) for(int i=((int)(n)-1);i>=m;i--)
#define all(x) (x).begin(),(x).end()
typedef long long ll;
const int MOD=1000000007;
using namespace std;
template<class T>
vector<T> make_vec(size_t a){
    return vector<T>(a);
}
template<class T, class... Ts>
auto make_vec(size_t a, Ts... ts){
    return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}

int main(){
    int n;
    cin>>n;
    vector<int> a(3*n);
    
    priority_queue<int,vector<int>,greater<int>> quel;
    priority_queue<int> quer;
    
    ll left=0,right=0;
    
    rep(i,0,3*n){
        cin>>a.at(i);
        if(i<n) {
            left+=a.at(i);
            quel.push(a.at(i));
        }
        else if(i>=2*n) {
            right+=a.at(i);
            quer.push(a.at(i));
        }
    }
    
    vector<ll> l(n+1);
    rep(i,n,2*n){
        l.at(i-n)=left;
        left+=a.at(i);
        quel.push(a.at(i));
        left-=quel.top();
        quel.pop();
    }
    l.at(n)=left;
    
    vector<ll> r(n+1);
    rrep(i,2*n,n){
        r.at(i+1-n)=right;
        right+=a.at(i);
        quer.push(a.at(i));
        right-=quer.top();
        quer.pop();
    }
    r.at(0)=right;
    
    ll ans=-1000000000000000;
    rep(i,0,n+1) ans=max(ans,l.at(i)-r.at(i));
    cout<<ans<<endl;
}