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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define ll long long int
#define ld long double
#define yorn(f) cout<<((f)?"Yes":"No")<<endl;
#define YORN(f) cout<<((f)?"YES":"NO")<<endl;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repi(i, n) for (int i = 1; i < (int)(n); i++)
#define all(x) (x).begin(), (x).end()
#define put(x) cout << x << endl;
#define println(x) cout << x << endl;
using namespace std;

void dump(vector<ll> v) {
    rep(i, v.size()) cout << v[i] << " ";
    cout << endl;
}



int main()
{
    int n, k;
    cin >> n >> k;
    vector<ll> x(n);
    rep(i, n) cin >> x[i];

    int stop = n - 1;
    rep(i, n) if (x[i] >= 0) {
        stop = i;
        break;
    }

    ll mi = 1e10;
    rep(i, stop + 1) {
        int end = i + k;
        if (end <= n) {
            mi = min(mi, 2 * abs(min(0LL, x[i])) + max(0LL, x[end-1]));
            mi = min(mi, abs(min(0LL, x[i])) + 2 * max(0LL, x[end-1]));
        }
    }
    
    put(mi);

    return 0;
}