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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int N, K;
    cin >> N >> K;

    int p = 0, n = 0, z = 0;
    vector<int> a(N);
    for (int& x : a) {
        cin >> x;
        if (x > 0) p++;
        else if (x < 0) n++;
        else z++;
    }

    int ans = 1 << 29;
    for (int i = 0; i < N; i++) {
        // i : point of reverse
        // first, go to dir placing a[i]
        int sub = 0, cnt = 0, l, r;
        if (z) l = r = n;
        else l = n - 1, r = n;
        
        int prev = 0;
        if (i < n) {
            for (int j = l; j >= i; j--) {
                // cerr << abs(prev - a[j]) << endl;
                sub += abs(prev - a[j]);
                prev = a[j];
        // cerr << i << " " << sub << " " << cnt << endl;
                if (++cnt >= K) goto ok;
            }
            for (int j = r; j < N; j++) {
                sub += abs(prev - a[j]);
                prev = a[j];
        // cerr << i << " " << sub << " " << cnt << endl;
                if (++cnt >= K) goto ok;
            }
        }
        else {
            for (int j = r; j <= i; j++) {
                sub += abs(prev - a[j]);
                prev = a[j];
        // cerr << i << " " << sub << " " << cnt << endl;
                if (++cnt >= K) goto ok;
            }
            for (int j = l; j >= 0; j--) {
                sub += abs(prev - a[j]);
                prev = a[j];
        // cerr << i << " " << sub << " " << cnt << endl;
                if (++cnt >= K) goto ok;
            }
        }
ok:
        // cerr << i << " " << sub << " " << cnt << endl;
        if (cnt == K) ans = min(ans, sub);
    }
    cout << ans << endl;

    return 0;
}