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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

const   fi='';//B.inp';
        fo='';//B.out';
var     f:text;
        i,j,n,top,top1:longint;
        kq:int64;
        heap,heap1,a,l,g:array[0..5000000] of longint;
(*===================================*)
procedure inp;
Begin
assign(f,fi);
reset(f);
readln(f,n);
for i:=1 to 3*n do read(f,a[i]);
close(f);
End;
(*===================================*)
procedure swap(var a,b:longint);
var y:longint;
Begin
y:=a;
a:=b;
b:=y;
End;
(*===================================*)
procedure upheap(i:longint);
Begin
if (i=1) or (heap[i]<heap[i div 2]) then exit;
swap(heap[i],heap[i div 2]);
upheap(i div 2);
End;
(*===================================*)
procedure downheap(i:longint);
var j:longint;
Begin
j:=i*2;
if j>top then exit;
if (j<top) and (heap[j]<heap[j+1]) then inc(j);
if heap[i]<heap[j] then
 begin
  swap(heap[i],heap[j]);
  downheap(j);
 end;
End;
(*===================================*)
procedure push(x:longint);
Begin
inc(top);
heap[top]:=x;
upheap(top);
End;
(*===================================*)
function pop:longint;
Begin
pop:=heap[1];
heap[1]:=heap[top];
dec(top);
downheap(1);
End;
(*===================================*)
procedure upheap1(i:longint);
Begin
if (i=1) or (heap1[i]>heap1[i div 2]) then exit;
swap(heap1[i],heap1[i div 2]);
upheap1(i div 2);
End;
(*===================================*)
procedure downheap1(i:longint);
var j:longint;
Begin
j:=i*2;
if j>top1 then exit;
if (j<top1) and (heap1[j]>heap1[j+1]) then inc(j);
if heap1[i]>heap1[j] then
 begin
  swap(heap1[i],heap1[j]);
  downheap1(j);
 end;
End;
(*===================================*)
procedure push1(x:longint);
Begin
inc(top1);
heap1[top1]:=x;
upheap1(top1);
End;
(*===================================*)
function pop1:longint;
Begin
pop1:=heap1[1];
heap1[1]:=heap1[top1];
dec(top1);
downheap1(1);
End;
(*===============================*)
procedure trong;
Begin
for i:=1 to n do begin l[i]:=l[i-1]+a[i]; push1(a[i]); end;
for i:=n+1 to 3*n do
  begin
   l[i]:=l[i-1]+a[i];
   push1(a[i]);
   l[i]:=l[i]-pop1;
  end;

for i:=3*n downto 2*n+1 do begin g[i]:=g[i+1]+a[i]; push(a[i]); end;
for i:=2*n downto 1 do
 begin
  g[i]:=g[i+1]+a[i];
  push(a[i]);
  g[i]:=g[i]-pop;
 end;
End;
(*===============================*)
BEGIN
inp;
assign(f,fo);
rewrite(f);
trong;
kq:=-maxlongint*100000;
for i:=n to 2*n do
 if l[i]-g[i+1]>kq then kq:=l[i]-g[i+1];
writeln(f,kq);
close(f);
END.