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>
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(x) (x).begin(),(x).end()
//#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
typedef long long ll; const int inf = INT_MAX / 2; const ll infl = 1LL << 60;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
//------------------------------------------------------------------------------
//------------------------------------------------------------------------------
void _main()
{
    int N;
    cin >> N;
    vector<ll> a(3 * N);
    ll sumf = 0;
    priority_queue<ll, vector<ll>, greater<ll>> qf;
    vector<pair<ll, int>> vb;
    rep(i, 0, 3 * N)
    {
        cin >> a[i];
        if (i < N)
        {
            sumf += a[i];
            qf.push(a[i]);
        }
        else
        {
            vb.push_back(make_pair(a[i], i));
        }
    }

    sort(vb.begin(), vb.end());

    ll sumb = 0;
    rep(i, 0, N) sumb += vb[i].first;

    ll ans = sumf - sumb;

    ll vbn = N;

    set<pair<ll, int>> s;

    rep(i, N, 2 * N)
    {
        if (a[i] > qf.top())
        {
            sumf -= qf.top();
            sumf += a[i];

            qf.pop();
            qf.push(a[i]);

            while (s.find(make_pair(vb[vbn].first, vb[vbn].second)) != s.end())
            {
                sumb -= vb[vbn].first;
                ++vbn;
                sumb += vb[vbn].first;
            }
        }

        const auto it = lower_bound(vb.begin(), vb.end(), make_pair(a[i], i));

        s.insert(make_pair(a[i], i));

        if (it - vb.begin() < vbn)
        {
            sumb -= a[i];
            sumb += vb[vbn].first;
            ++vbn;
        }

        chmax(ans, sumf - sumb);
    }

    cout << ans << endl;
}