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<iomanip>
#include<algorithm>
#include<cmath>
#include<vector>
#include<string>
#include<bitset>
#include<queue>
#include<set>
#include<map>
using namespace std;

#define FOR(i,a,b) for (int i=(a);i<(b);++i)
#define ROF(i,b,a) for (int i=(b);i>(a);--i)
#define REP(i,n) FOR(i,0,n)
#define PER(i,n) ROF(i,n-1,-1)

typedef int i32;
typedef long long int i64;
typedef unsigned long long int ui64;
typedef float f32;
typedef double f64;
typedef long double f128;
typedef pair<int,int> Pi;

constexpr int INF = 1e9 + 7;
constexpr int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1};

int main() {
    int N,K; cin>>N>>K;
    i64 x[N]; REP(i,N) cin>>x[i];
    i64 ans = 1e14;
    i64 xm[N], xp[N];
    REP(i,N) {
        xm[i] = x[i]-x[0];
        xp[i] = x[N-1]-x[N-1-i];
    }
    REP(i,N-K+1) {
        ans = min({xm[i+K-1]-xm[i]+abs(x[i]),xp[i+K-1]-xp[i]+abs(x[i]),ans});
    }
    cout<<ans<<endl;
}