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 PyPy3 (2.4.0) (judge verdict: Accepted)

import bisect
import collections
import sys
import heapq

sys.setrecursionlimit(100000)
input = sys.stdin.readline
ACMOD = 1000000007
INF = 1 << 32


def lmi():
    return list(map(int, input().split()))


def llmi(n):
    return [lmi() for _ in range(n)]


N = int(input())
A = lmi()

lq = []  # normal
hq = []  # * -1
for a in A[:N]:
    heapq.heappush(lq, a)
for a in A[2 * N:]:
    heapq.heappush(hq, - a)

dpl = [0 for _ in range(N + 1)]  # middle をNまで消費したときの (a′の前半N要素の総和)
dph = [0 for _ in range(N + 1)]  # middle をNまで消費したときの (a′の後半N要素の総和)

middle = A[N:2 * N]
dpl[0] = sum(A[:N])
last_val = dpl[0]
for count in range(1, N + 1):
    heapq.heappush(lq, middle[count - 1])
    v = heapq.heappop(lq)
    dpl[count] = dpl[count - 1] - v + middle[count - 1]

dph[N] = sum(A[2 * N:])
for count in reversed(range(N)):
    heapq.heappush(hq, -middle[count])
    v = -  heapq.heappop(hq)  # bigger value
    dph[count] = dph[count + 1] - v + middle[count]


print(max(dpl[i] - dph[i] for i in range(N+1)))