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: Wrong Answer)

//g++ -std=c++14 test.cpp -o test.out

#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <cassert>
#include <algorithm>
#include <functional>
#include <iostream>
#include <iomanip>
#include <vector>
#include <queue>
#include <map> 
#include <utility>
#include <limits.h>
#include <bitset>
#include <set>
using namespace std;
 
#define LL long long int
const LL INF = (LL)(1e18)+1;
const int INF_INT = 2147483647-1e6-1;
const LL mod = 1000000007ll;
const int mod_int = 1000000007;

LL N,K;
LL x[100000];

LL ans = INF;

void solve(){
    //それぞれ座標0から見て、正の方向と負の方向にあるろうそくを格納
    //ただし0は正とみなす
    vector<LL> positive,negative;
    for(int i=0;i<N;i++){
        if(x[i]<0) negative.push_back(-x[i]);
        else positive.push_back(x[i]);
    }
    reverse(negative.begin(),negative.end());

    //正の方向に行ったっきりな場合
    if(positive.size()>=K) ans = positive[K-1];
    //負の方向に行ったっきりな場合
    if(negative.size()>=K) ans = min(ans,negative[K-1]);

    if(positive.size()==0 || negative.size()==0 || K==1) return;

    LL p = min(K,(LL)positive.size());//正の方向にある、調べる予定のろうそくの数
    LL n = K-p;//負の方向にある、調べる予定のろうそくの数

    while(1){
        if(p<=0 || n>=negative.size()) break;
        //正の方向に行ってから負の方向に引き返す場合
        ans = min(ans,positive[p-1]*2 + negative[n-1]);
        //負の方向に行ってから正の方向に引き返す場合
        ans = min(ans,negative[n-1]*2 + positive[p-1]);
        p--;
        n++;
    }
}

int main(){
    cin >> N >> K;
    for(int i=0;i<N;i++){
        cin >> x[i];
    }
 
    solve();

    cout << ans << endl;
    return 0;
}