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 <iostream>
#include <queue>
#include <stack>
#include <vector>
#include <cstring>
#include <cmath>
#include <map>
#include <unordered_map>
#include <set>
#include <algorithm>
#include <iomanip>
#define F first
#define S second
#define PB push_back
#define PF push_front
#define MP make_pair
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
const int maxn = 1e5 + 10;
const int maxm = 1e5 + 10;
const int mod = 1e9 + 7;
const int inf = 1e9;

vector <int> v, g;

int main() {
    ios_base::sync_with_stdio(false);
    int n, k;
    cin >> n >> k;
    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        if (x < 0)
            v.PB (x);
        else
            g.PB (x);
    }
    reverse (v.begin(), v.end());
    int ans = inf;
    for (int i = 0; i < min ((int)g.size(), k); i++) {
        if (k - i - 2 < v.size() and k - i - 2 >= 0) {
            ans = min (ans, 2 * g[i] - v[k - i - 2]);
            ans = min (ans, -2 * v[k - i - 2] + g[i]);
        }
    }
    if (v.size() >= k)
        ans = min (ans, v[k - 1]);
    if (g.size() >= k)
        ans = min (ans, g[k - 1]);
    cout << ans << endl;
}