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 rep(i,n) for (int i = 0; i < n; ++i)
#define rep2(i,s,n) for (int i = s; i < n; ++i)
#define all(a) a.begin(),a.end()
#define tmax(a,b,c) max(a,max(b,c))
#define tmin(a,b,c) min(a,min(b,c))
#define pb push_back
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
template<class T> inline bool chmin(T& a,T b) {if(a > b){a = b; return true;} return false;}
template<class T> inline bool chmax(T& a,T b) {if(a < b){a = b; return true;} return false;}
const int inf = 1001001001;

int main() {
    int n;
    cin >> n;
    vector<ll> a(n*3),mx(n*3),mi(n*3);
    rep(i,n*3) cin >> a[i];
    priority_queue<ll> nmi;
    ll sum = 0;
    for(int i = n*3-1; i >= n*2; i--) {
        nmi.push(a[i]);
        sum += a[i];
    }
    mi[n*2] = sum;
    for(int i = n*2-1; i >= n; i--) {
        sum += a[i];
        nmi.push(a[i]);
        sum -= nmi.top();
        nmi.pop();
        mi[i] = sum;
    }
    priority_queue<ll,vector<ll>,greater<>> nmx;
    sum = 0;
    rep(i,n) {
        nmx.push(a[i]);
        sum += a[i];
    }
    mx[n-1] = sum;
    rep2(i,n,n*2) {
        sum += a[i];
        nmx.push(a[i]);
        sum -= nmx.top();
        nmx.pop();
        mx[i] = sum;
    }
    ll ans = -1001001001001001;
    rep2(i,n-1,n*2) {
        chmax(ans,mx[i]-mi[i+1]);
    }
    cout << ans << endl;
}