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 rep(i,n) for (int i = 0; i < (int)(n); i++)
using namespace std;
using ll = long long;
using vint = vector<int>;
using vvint = vector<vector<int>>;
using vll = vector<ll>;
const ll inf = 1e18;

int main() {
    int n, k;
    cin >> n >> k;
    vint x(n);
    rep(i,n) cin >> x[i];
    vll tmp;
    int border = lower_bound(x.begin(),x.end(),0) - x.begin();
    for (int i = -k; i <= k-1; i++) {
        if (border+i >= 0 && border+i < n) tmp.push_back(x[border+i]);
    }
    ll ans = inf;
    for (int i = 0; i <= tmp.size()-k; i++) {
        if (tmp[i+k-1] < 0) ans = min(-tmp[i],ans);
        else if (tmp[i] >= 0) ans = min(tmp[i+k-1],ans);
        else ans = min(tmp[i+k-1]-tmp[i]+min(tmp[i+k-1],-tmp[i]),ans);
    }
    cout << ans << endl;
}