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;
#define rep(i,n) for(ll i=0; i<n; i++)
typedef long long ll;
typedef long long unsigned int llu;
ll mod = 1000000007;
ll INF = 1000000009;

/*
一番近いろうそくに火を付ける？
-1 100 101 102
-101 -100 -1 99 100

一番狭い区間に火を付ける？
10 13 16 100 101 102

いずれにせよ、連続したk個のろうそくに火を付けるので、
連続したk個のろうそくに火を付ける為に必要なコストを計算する
区間が0を含んでいなければ、最も遠いろうそくまでの距離
0を含んでいれば、[-p,0][0,q]とし、
min(2*p+q,p+2*q)
*/

void solve(){
    ll n;
    ll k;
    cin >> n >> k;
    ll x[n+10];
    rep(i,n){
        cin >> x[i];
    }
    ll ans = INF;
    for(ll i=k-1; i<n; i++){
        ll cost=0;
        //区間 x[i-k+1], x[i-k+2], ..., x[i]
        if(x[i]<=0){
            cost = -x[i-k+1];
        }
        else if(x[i-k+1]>=0){
            cost = x[i];
        }
        else{
            ll p=-x[i-k+1];
            ll q=x[i];
            cost = min(2*p+q, p+2*q);
        }
        ans = min(ans, cost);
    }
    cout << ans << endl;
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  solve();
  return 0;
}