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<vector>
#include<algorithm>
#include<string>
#include<set>
#include<map>
#include<queue>
#define REP(i,a) for (int i = 0;i < (a);++i)
#define FOR(i,a,b) for (int i = (a);i < (b); ++i)
#define FORR(i,a,b) for (int i = (a);i >= (b); --i)
#define ALL(obj) (obj).begin,(obj).end()
#define SORT(list) sort(ALL((list)));
#define MOD 1000000007
using namespace std;
using ll = long long;
int main(){
    int n,k;
    cin >> n >> k;
    vector<ll> x(n);
    REP(i,n)cin >> x[i];
    ll answer = 1000000000;
    int l=0;
    if(k  == 1){
        REP(i,n)answer = min(answer,llabs(x[i]));
        cout << answer << endl;
        return 0;
    }
    
    int r = k-1;
    while(r < n-1){
        answer = min(answer,llabs(x[l])+llabs(x[r]-x[l]));
        answer = min(answer,llabs(x[r])+llabs(x[r]-x[l]));
        l++;
        r++;
    }
    cout << answer << endl;
    return 0;
}