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>
using namespace std;
 
#define REP(i,n) for(int i=0; i < (n); ++i)
#define REPR(i,n) for(int i=(n); i >=0; --i)
#define FOR(i, m, n) for(int i = (m); i < (n); ++i)
 
typedef long long ll;
#define INF 1e9
 
 
int main(){
  int N, K, zero = 1e9;
  cin >> N >> K;
  vector<int> x(N), minus, plus(1,0);
  REP(i, N){
    cin >> x[i];
    if(x[i] > 0){
      plus.push_back(x[i]);
    }
    else if(x[i] < 0){
      minus.push_back(-x[i]);
    }
    else{
      K--;
    }
  }
  minus.push_back(0);
  int ans = 1e9;
  
  for(int i =max((int)(K-minus.size()),0); i <= K && i < plus.size(); i++){
    // cout << "minus.size():" << minus.size() << " K:" << K << " i:" << i << endl;
    // cout << minus[minus.size()-1-K+i] << endl;
    ans = min(ans, (2*plus[i]+(minus[minus.size()-1-K+i])));
    ans = min(ans, (plus[i] + 2*(minus[minus.size()-1-K+i])));
  }
  cout << ans << endl;


 
  return 0;
}