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 <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define erep(i,a,b) for(int i=a;i<=(int)(b);++i)
#define per(i,a,b) for(int i=(a);i>(b);--i)
#define eper(i,a,b) for(int i=(a);i>=b;--i)
#define pb push_back
#define mp make_pair
#define INF (1<<30)-1
#define MOD 1000000007
#define all(x) (x).begin(),(x).end()
#define vii vector<int>
#define vll vector<long long>
using namespace std;
typedef long long ll;
typedef pair<int,int> Pii;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a/gcd(a, b)*b;}

int n, k;
ll ans = INF;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
    cin >> n >> k;
    vll x(n), sum(n);
    rep(i, 0, n) cin >> x[i];
    rep(i, 0, n-k+1) {
        ans = min(ans, min(abs(x[i]) + abs(x[i] - x[i + k - 1]), abs(x[i + k - 1] + abs(x[i + k - 1] - x[i]))));
    }
    cout << ans << endl;
    return 0;
}