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(int (i)=0;(i)<(n);(i)++)
#define repp(i,m,n) for(int (i)=(m);(i)<(n);(i)++)
#define repm(i,n) for(int (i)=(n-1);(i)>=0;(i)--)
#define INF (1ll<<60)
#define all(x) (x).begin(),(x).end()
typedef long long lint;
const lint MOD =1000000007;
const lint MAX = 10000000;
using Graph =vector<vector<lint>>;
typedef pair<lint,lint> P;

 
lint fac[MAX], finv[MAX], inv[MAX];

void COMinit() // 二項係数を求める時は前処理としてCOMinit()をする
{
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (lint i = 2; i < MAX; i++)
    {
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
// 二項係数計算
long long COM(lint n, lint k)
{
    if (n < k)
        return 0;
    if (n < 0 || k < 0)
        return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

lint primary(lint num)
{
    if (num < 2) return 0;
    else if (num == 2) return 1;
    else if (num % 2 == 0) return 0; // 偶数はあらかじめ除く

    double sqrtNum = sqrt(num);
    for (int i = 3; i <= sqrtNum; i += 2)
    {
        if (num % i == 0)
        {
            // 素数ではない
            return 0;
        }
    }

    // 素数である
    return 1;
}
lint lcm(lint a,lint b){
      return a/__gcd(a,b)*b;
  }
 
  int main(){
 lint N,K;
 cin>>N>>K;
 vector<lint>minus;
 vector<lint>plus;
 lint ans=INF;
 rep(i,N){
     lint v;
     cin>>v;
     if(v<0)minus.push_back(v);
     else plus.push_back(v);
 }
 reverse(all(minus));
 rep(i,min(K,(lint)minus.size())){
     lint accum=0;
     accum+=abs(minus[i])*2;
     if(K-i-1<=plus.size()&&i!=(K-1)){
         accum+=plus[K-i-2];
     }else if(i!=(K-1))continue;
     if(i==(K-1))accum-=abs(minus[i]);
        ans=min(ans,accum);
 }

  rep(i,min(K,(lint)plus.size())){
     lint accum1=0;
     accum1+=abs(plus[i])*2;
     if(K-i-1<=minus.size()&&i!=(K-1)){
         accum1-=minus[K-i-2];
     }else if(i!=(K-1)) continue;
          if(i==(K-1))accum1-=abs(plus[i]);
        ans=min(ans,accum1);
 }
    cout<<ans;
}