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) for(ll i=0;i<n;++i)
#define all_map(itr,mp) for(auto itr=mp.begin();itr!=mp.end();itr++)
#define ALL(a) (a).begin(),(a).end()

using namespace std;
using ll  = long long;
using pll = pair<ll, ll>;

const ll MOD  = 1e9+7;
const ll LINF = 1LL << 60;
const int INF = 1e9 + 7;

int main(){
    ll n;
    cin >> n;

    vector<ll> a(3*n);
    
    rep(i, 3*n)cin >> a[i];

    vector<ll> plus(3*n, 0);
    priority_queue<ll, vector<ll>, greater<ll>> plus_que;
    rep(i, n){
        plus[i+1] = plus[i] + a[i];
        plus_que.push(a[i]);
    }

    for(ll i = n; i < 2*n; ++i){
        ll top = plus_que.top();
        if(a[i] > top){
            plus[i+1] = plus[i] - top + a[i];
            plus_que.pop();
            plus_que.push(a[i]);
        }
        else plus[i+1] = plus[i];
    }


    vector<ll> minus(3*n, 0);
    priority_queue<ll> minus_que;
    rep(i, n){
        minus[i+1] = minus[i] + a[3*n-1-i];
        minus_que.push(a[3*n-1-i]);
    }
    
    for(ll i = n; i < 2*n; ++i){
        ll top = minus_que.top();
        if(a[3*n-1-i] < top){
            minus[i+1] = minus[i] - top + a[3*n-1-i];
            minus_que.pop();
            minus_que.push(a[3*n-1-i]);
        }
        else minus[i+1] = minus[i];
    }

    ll ans = -1*LINF;
    for(ll i = n; i <= 2*n; ++i){
        ans = max(ans, plus[i] - minus[3*n-i]);
    }

    cout << ans << endl;
}