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 ll long long
#define M 1000000007
#define pll pair<ll,ll>
#define rep(i,a,b) for(ll i=a;i<b;i++)
#define mll map<ll,ll>
#define vl vector<ll>
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(),a.end()
#define F first
#define S second
using namespace std;
int main()
{
    ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
    ll n;
    cin>>n;
    ll a[3*n],l[n+1],r[n+1];
    rep(i,0,3*n)
    cin>>a[i];
    priority_queue<ll,vector<ll>,greater<ll>>pq;
    priority_queue<ll>pq2;
    ll sum=0,ans=LLONG_MIN;
    rep(i,0,2*n)
    {
        sum+=a[i];
        pq.push(a[i]);
        if(pq.size()>n)
        {
            sum-=pq.top();
            pq.pop();
        }
        if(pq.size()==n)
        l[i-n+1]=sum;
    }
    sum=0;
    for(ll i=3*n-1;i>=n;i--)
    {
        sum+=a[i];
        pq2.push(a[i]);
        if(pq2.size()>n)
        {
            sum-=pq2.top();
            pq2.pop();
        }
        if(pq2.size()==n)
        r[2*n-i]=sum;
    }
    rep(i,0,n+1)
    ans=max(ans,l[i]-r[n-i]);
    cout<<ans;
}