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 Julia (0.5.0) (judge verdict: Wrong Answer)

parseInt(x) = parse(Int, x)
parseMap(x::Array{SubString{String},1}) = map(parseInt, x)
function heap(a::Array{Int,1})
	h = [a[1]]
	for i in 2:length(a)
		hpush(h,a[i])
	end
	h
end

function hpush(h::Array{Int,1},x)
	push!(h,x)
	l = length(h)
	if l>1
		k = div(l,2)
		while h[k]>x&&k>=1
			tmp = h[k]
			h[k] = x
			h[l] = tmp
			k = max(div(k,2),1)
			l = div(l,2)
		end
	end
end

function hpop(h::Array{Int,1})
	x = h[1]
	z = pop!(h)
	if !isempty(h)
		h[1] = z
		k = 1
		l = length(h)
		while k <= l
			if 2*k>=l
				break
			elseif 2*k+1>=l
				if h[2*k]>=z
					break
				else
					tmp = h[2*k]
					h[2*k] = z
					h[k] = tmp
					k = 2*k
				end
			else
				if h[2*k]>=z&&h[2*k+1]>=z
					break
				else
					t = h[2*k]<h[2*k+1]?2*k:2*k+1
					tmp = h[t]
					h[t] = z
					h[k] = tmp
					k = t
				end
			end
		end
	end
	x
end

function heapl(a::Array{Int,1})
	h = [a[1]]
	for i in 2:length(a)
		hpushl(h,a[i])
	end
	h
end

function hpushl(h::Array{Int,1},x)
	push!(h,x)
	l = length(h)
	if l>1
		k = div(l,2)
		while h[k]<x&&k>=1
			tmp = h[k]
			h[k] = x
			h[l] = tmp
			k = max(div(k,2),1)
			l = div(l,2)
		end
	end
end

function hpopl(h::Array{Int,1})
	x = h[1]
	z = pop!(h)
	if length(h)>0
		h[1] = z
		k = 1
		l = length(h)
		while k <= l
			if 2*k>l
				break
			elseif 2*k+1>l
				if h[2*k]<=z
					break
				else
					tmp = h[2*k]
					h[2*k] = z
					h[k] = tmp
					k = 2*k
				end
			else
				if h[2*k]<=z&&h[2*k+1]<=z
					break
				else
					t = h[2*k]<h[2*k+1]?2*k+1:2*k
					tmp = h[t]
					h[t] = z
					h[k] = tmp
					k = t
				end
			end
		end
	end
	x
end

function main()
	n = readline() |> parseInt
	a = readline() |> split |> parseMap
	b = heap(a[1:n])
	c = heapl(a[2n+1:3n])
	l = zeros(Int,n+1)
	r = zeros(Int,n+1)
	l[1] = sum(b)
	r[n+1] = sum(c)
	for i in 1:n
		if b[1] < a[n+i]
			p=hpop(b)
			hpush(b,a[n+i])
			l[i+1] = l[i]-p+a[n+i]
		else
			l[i+1] = l[i]
		end
		if c[1] > a[2n+1-i]
			q = hpopl(c)
			hpushl(c,a[2n+1-i])
			r[n+1-i] = r[n+2-i]-q+a[2n+1-i]
		else
			r[n+1-i] = r[n+2-i]
		end
	end
	s = 0
	for i in 1:n+1
		s = max(s,l[i]-r[i])
	end
	println(s)
end

main()