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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<vector>
#include<stdlib.h>
#include<time.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<map>
using namespace std;
int N;
int check=0;

long long int Min(long long int a, long long int b){
 if(a<=b)return a;
 else return b;
}

long long int Abs(long long int a){
  if(a>0)return a; 
  else return a*-1;
}

int main(void){
 int n,k;
 int x;
 int a;
 long long int ans=300000000000;
 vector<int> A;
 vector<int> B;
 

 cin>>n>>k;

 for(int i=0; i<n; i++){
  cin>>a;
  if(a<0)A.push_back(Abs(a));  
  else B.push_back(a);
 }
 
 sort(A.begin(),A.end());
 sort(B.begin(),B.end());
 
 if(A.size()>=k)ans=Min(ans,A[k-1]);
 if(B.size()>=k)ans=Min(ans,B[k-1]);
 //for(int i=0; i<A.size();i++)std::cout<<A[i]<<std::endl;

 for(int i=0; i<A.size(); i++){
  long long int sum=30000000000;
  if(k-1-(i+1)<B.size())sum=(long long int)A[i]*2+B[k-1-(i+1)];
  ans=Min(ans,sum);
  if(k-1-(i+1)<B.size())sum=A[i]+(long long int) B[k-1-(i+1)]*2;
  ans=Min(ans,sum);
 }
 std::cout<<ans<<std::endl;
}