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>
//#pragma GCC optimize("O3")
//#pragma GCC target ("sse4")
using namespace std;
#define PB push_back
#define all(x) (x).begin(),(x).end()
#define MEM(x,a) memset((x),a,sizeof((x)))
#define F first
#define S second
#define imx INT_MAX
#define inI(x) scanf("%d",&x)
#define inI2(x,y) scanf("%d %d",&x,&y)
#define inI3(x,y,z) scanf("%d %d %d",&x,&y,&z)
const long long MOD = (long long)(1e9+7);
const long long MMX = (long long)(1e18);
typedef long long LL;
int n,num[300005];
LL ss,mx[300005],mn[300005],ans;
priority_queue<int>pq;
int main()
{
    ans=-MMX;
    scanf("%d",&n);
    for(int i=1;i<=3*n;i++)
    {
        scanf("%d",&num[i]);
    }
    for(int i=1;i<=n;i++)
    {
        pq.push(-num[i]);
        ss+=num[i];
        mx[i]=ss;
    }
    for(int i=n+1;i<=3*n;i++)
    {
        if(-pq.top()<num[i])
        {
            ss+=pq.top();
            pq.pop();
            ss+=num[i];
            pq.push(-num[i]);
        }
        mx[i]=ss;
    }
    while(!pq.empty())pq.pop();
    ss=0;
    for(int i=n*3;i>n*2;i--)
    {
        pq.push(num[i]);
        ss-=num[i];
        mn[i]=ss;
    }
    for(int i=(n*2);i>0;i--)
    {
        if(pq.top()>num[i])
        {
            ss+=pq.top();
            pq.pop();
            ss-=num[i];
            pq.push(num[i]);
        }
        mn[i]=ss;
    }
    for(int i=n;i<=n*2;i++)
    {
        //printf("%d %lld %lld\n",i,mx[i],mn[i+1]);
        ans=max(ans,mx[i]+mn[i+1]);
    }
    printf("%lld",ans);
}