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: Time Limit Exceeded)

#include <bits/stdc++.h>
#include <cstdio>
#include <cstdlib>

#ifdef _MSC_VER
#include <Windows.h>
#else
#include <unistd.h>
#endif

using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

//math
//-------------------------------------------
template<class T> inline T sqr(T x) {return x*x;}

//typedef
//------------------------------------------
typedef pair<int, int> PII;
typedef pair<long, long> PLL;
typedef long long LL;
//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB emplace_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SORT(c) sort((c).begin(),(c).end())

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI  = acos(-1.0);
const LL MOD=(double)1e9+7;
//clear memory
#define CLR(a) memset((a), 0 ,sizeof(a))

#define DEBUG
long a[300003];

int main(){
    long N;
    cin>>N;
    REP(i,3*N){
        cin>>a[i];
    }
    LL sum1=0LL,sum2=0LL;
    LL res=-LONG_LONG_MAX;
    for(long p=N;p<=2L*N;p++){//どこで分割するか？Nから2Nまでの間に分割点がある
        priority_queue<long, vector<long>, greater<long> > minpq1;
        priority_queue <long> maxpq2;
        sum1=0;sum2=0;
        for(long i=p;i<3*N;i++){
            maxpq2.push(a[i]);
            sum2+=a[i];
        }
        while(SZ(maxpq2)>N){
            sum2-=maxpq2.top();
            maxpq2.pop();
        }
        for(long i=0;i<p;i++){
            minpq1.push(a[i]);
            sum1+=a[i];
        }
        while(SZ(minpq1)>N){
            sum1-=minpq1.top();
            minpq1.pop();
        }
        res=max(res,sum1-sum2);
    }
    //cout<<rp<<endl;
    /*for(long i=rp+1;i<2*N;i++){
        maxpq2.push(a[i]);
        sum2+=a[i];
    }
    while(SZ(maxpq2)>N){
        sum2-=maxpq2.top();
        maxpq2.pop();
    }*/
    //cout<<sum1<<" "<<sum2<<endl;
    cout<<res<<endl;
    return 0;
}