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>
using namespace std;
 
#define pb push_back
#define ll long long
#define maxn 300005
#define fr(i,j,k) for(int i=j;i<k;i++)
#define f(n) fr(i,0,n)
#define f1(n) fr(i,1,n+1)
#define ms(i) memset(i,0,sizeof(i));
#define ms1(i) memset(i,-1,sizeof(i));
#define F first
#define S second
#define all(x) x.begin(), x.end()
const int mod = 998244353;

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    vector<ll>v(n * 3);
    ll pre[n + 5] = {};
    ll suf[n + 5] = {};
    for (auto &i : v) {
        cin >> i;
    }
    priority_queue<int,vector<int>,greater<int>>pq;
    ll sum = 0;
    for (int i = 0 ; i < n ; i++) {
        pq.push(v[i]);
        sum += v[i];
    }
    for (int i = 0 ; i <= n ; i++) {
        pre[i] = sum;
        if (v[i + n] > pq.top()) {
            sum -= pq.top();
            pq.pop();
            sum += v[i + n];
            pq.push(v[i + n]);
        }
    }
    priority_queue<int>pq2;
    sum = 0;
    for (int i = n * 2 ; i < n * 3 ; i++) {
        pq2.push(v[i]);
        sum += v[i];
    }
    suf[n] = sum;
    for (int i = n - 1 ; i >= 0 ; i--) {
        if (v[i + n] < pq2.top()) {
            sum -= pq2.top();
            pq2.pop();
            sum += v[i + n];
            pq2.push(v[i + n]);
        }
        suf[i] = sum;
    }
    ll mx = -1e18;
    for (int i = 0 ; i <= n ;i++) {
        mx = max(mx, pre[i] - suf[i]);
    }
    cout << mx << '\n';
}