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>
#define ll long long
using namespace std;
const int N = 4e5 + 5, mod = 998244353;


int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int n, k;
    cin >> n >> k;
    vector<int> v, v2;
    for(int i = 0; i < n; i++){
        int x;
        cin >> x;
        if(x < 0) v2.push_back(-x);
        else v.push_back(x);
    }
    int ans = 2e9;
    sort(v2.begin(), v2.end());
    for(int i = 0; i < v.size(); i++){
        int idx = k - (i + 1) - 1;
        if(idx < v2.size()) ans = min(ans, v[i] + v[i] + v2[idx]);
    }
    for(int i = 0; i < v2.size(); i++){
        int idx = k - (i + 1) - 1;
        // cout << idx << ' ' << v[idx] << '\n';
        if(idx < v.size()) ans = min(ans, v2[i] + v2[i] + v[idx]);
    }
    if(v.size() >= k) ans = min(ans, v[k - 1]);
    if(v2.size() >= k) ans = min(ans, v2[k - 1]);
    cout << ans;


    return 0;

}