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 <bits/stdc++.h>
using namespace std;
using ll=long long;
const ll MOD=1000000007;
const double PI=3.14159265358979;
const ll INF= pow(10,18);
typedef pair<ll,ll> P;
typedef vector<ll> vl;
typedef vector<vl> vvl;
#define FOR(i,a,b) for(ll i=a;i<b;i++)
#define rep(i,n) FOR(i,0,n)
string abc="abcdefghijklmnopqrstuvwxyz";
string ABC="ABCDEFGHIJKLMNOPQRSTUVWXYZ";

int main(){
  ll n,k;
  cin >> n >> k;
  vl x(n);
  ll a=-1;
  rep(i,n){
    cin >> x[i];
    if(x[i]<0){
      a=i;
    }
  }
  vl dp_minus(n+2,INF);
  vl dp_plus(n+2,INF);
  dp_minus[0]=0;
  dp_plus[0]=0;
  for(ll i=a;i>=0;i--){
    dp_minus[a-i+1]=(-1)*x[i];
  }
  FOR(i,a+1,n){
    dp_plus[i-a]=x[i];
  }
  ll ans=INF;
  rep(i,k+1){
    ans=min(ans,min(2*dp_plus[i]+dp_minus[k-i],2*dp_minus[i]+dp_plus[k-i]));
    //cout << dp_plus[i] << " " << dp_minus[i] << endl;
  }
  cout << ans << endl;
}