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)

#include<bits/stdc++.h>

using namespace std;

#define allof(x) (x).begin(), (x).end()

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

int N, K;
vector<int> L, R;

int main(){
    cin.tie(0);
    cin.sync_with_stdio(0);
    
    cin >> N >> K;
    while(N--){
        int x;
        cin >> x;
        if(x <= 0){
            L.push_back(-x);
        } else {
            R.push_back(x);
        }
    }

    reverse(allof(L));

    int ans = 2e9;
    if((int)L.size() >= K){
        ans = L[K - 1];
    }

    if((int)R.size() >= K){
        ans = R[K - 1];
    }

    for(int a = 0; a < (int)L.size() && a < K - 1; a++){
        if(K - a - 2 < (int)R.size()){
            ans = min(ans, L[a] * 2 + R[K - a - 2]);
        }
    }

    cout << ans;
}