Problem:
Score : 300 points

Problem Statement

There are N candles placed on a number line.
The i-th candle from the left is placed on coordinate x_i.
Here, x_1 < x_2 < ... < x_N holds.

Initially, no candles are burning.
Snuke decides to light K of the N candles.

Now, he is at coordinate 0.
He can move left and right along the line with speed 1.
He can also light a candle when he is at the same position as the candle, in negligible time.

Find the minimum time required to light K candles.

Constraints

1 \leq N \leq 10^5

1 \leq K \leq N

x_i is an integer.

|x_i| \leq 10^8

x_1 < x_2 < ... < x_N

Input

Input is given from Standard Input in the following format:

N K
x_1 x_2 ... x_N

Output

Print the minimum time required to light K candles.

Sample Input 1

5 3
-30 -10 10 20 50

Sample Output 1

40

He should move and light candles as follows:

Move from coordinate 0 to -10.

Light the second candle from the left.

Move from coordinate -10 to 10.

Light the third candle from the left.

Move from coordinate 10 to 20.

Light the fourth candle from the left.

Sample Input 2

3 2
10 20 30

Sample Output 2

20

Sample Input 3

1 1
0

Sample Output 3

0

There may be a candle placed at coordinate 0.

Sample Input 4

8 5
-9 -7 -4 -3 1 2 3 4

Sample Output 4

10

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

var
	s:int64;
	n,k,i,x,y,t,a,p:longint;
	f,f1:array[-200010..200010] of longint;
function min(a,b:longint):longint;
begin
	if a<b then exit(a) else exit(b);
end;
begin
	read(n,k);
	x:=0; y:=0;
	t:=1; p:=0;
	fillchar(f1,sizeof(f1),$7f);
	for i:=1 to n do
	begin 
		read(a);
		if a>0 then inc(x);
		if a<0 then inc(y);
		if a=0 then dec(k);
		if (a>0) and (f[i-1]<=0) then t:=i;
		if a<>0 then
		begin 
			inc(p);
			f[p]:=a;
		end;
	end;
	if x=0 then
	begin
		writeln(-f[p-k+1]);
		halt;
	end;
	if y=0 then
	begin
		writeln(f[k]);
		halt;
	end;
	s:=$7f;
	for i:=1 to p do f1[i-t]:=abs(f[i]);
	y:=0-y;
	if x>=k then s:=min(s,f1[k-1]);
	if 0-y>=k then s:=min(s,f1[0-k]);
	for i:=-min(-y,k) to -1 do
		if i*(k+i-1)<=0 then
			s:=min(s,f1[i]*2+f1[k+i-1]);
	for i:=1 to min(x-1,k-1) do
		if i*(k-i-1)>=0 then
			s:=min(s,f1[i]*2+f1[0-(k-i-1)]);
	writeln(s);
end.