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)

/******Old School******/
#include<bits/stdc++.h>
using namespace std;
#define lld               long long int
#define ld                long double
#define pb                push_back
#define ppb               pop_back
#define mp                make_pair
#define nl                '\n'
#define all(c)            (c).begin(),(c).end()
#define rall(c)           (c).rbegin(),(c).rend()
#define sz(c)             (c).size()
#define tr(x,a)           for(auto &a : x)
#define psnt(x,a)         (x).find(a)!=(x).end()
#define vpsnt(x,a)        find(all(x),a)!=(x).end()
#define lexichar(x,y)     lexicographical_compare(x,x+strlen(x),y,y+strlen(y)) //modify the ranges
#define lexistr(x,y)      lexicographical_compare(x.begin(),x.end(),y.begin(),y.end()) 
#define MOD               1000000007
#define pi                3.1415926536
#define tod               1e18
#define itr(i,a,b)        for(lld i=a;i<=b;i++)
#define itrn(i,a,b)       for(lld i=a;i>=b;i--)
#define iot(n)            for(lld i=0;i<n;i++)
#define pls(n,arr)        lld arr[n]; iot(n)  cin>>arr[i];
 
typedef pair<lld,lld>     pii;
typedef pair<string,lld>  psi;
template <typename T>
bool mycomp(T x,T y){ 
   return (x==y);  //give your condition here
}
bool paircomp(const pair<lld,lld> &x,const pair<lld,lld> &y){ 
return x.second<y.second;
}
 
int main(){
       ios_base::sync_with_stdio(false);
             cin.tie(0);
            cout.tie(0);
     lld n; cin>>n;
    vector<lld> v(300005);
iot(3*n)   cin>>v[i];
lld csum=0,l[300005],r[300005];
  priority_queue<lld> pq; //max heap
priority_queue<lld,vector<lld>,greater<lld> > pq2;// min  heap
iot(3*n){
csum+=v[i];
pq2.push(v[i]);
if(pq2.size()>n){
csum-=pq2.top();
pq2.pop();
}
if(pq2.size()==n)  l[i]=csum;
else l[i]=-tod;
}
csum=0;
itrn(i,3*n-1,0){
pq.push(v[i]);
csum+=v[i];
if(pq.size()>n){
csum-=pq.top();
pq.pop();
}
if(pq.size()==n) r[i]=csum;
else  r[i]=tod;
}
lld ans=-tod;
iot(3*n-1){
ans=max(ans,l[i]-r[i+1]);
}
cout<<ans;
}