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>
using namespace std;

using ll = long long;
#define int long long

#define LOG(variable) cout << #variable":\t" << (variable) << endl
#define LOGCON(i, container) for(int (i) = 0; (i) < (container).size(); ++(i)) cout << (i) << ":\t" << (container)[(i)] << endl
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define REPS(i, r, n) for (int i = (r); i < (n); ++i)
#define REPR(i, n) for(int i = (n); i >= 0; --i) // from n to 0
#define REPRS(i, n, r) for(int i = (n); i >= (r); --i) // from n to r
#define REPOBJ(itr, obj) for(auto itr = (obj).begin(); itr != (obj).end() ; ++itr)
#define REPROBJ(itr, obj) for(auto itr = (obj).rbegin(), e = (obj).rend(); itr != e; ++itr)
#define COUTB(x) cout << (x) << "\n"
#define COUTS(x) cout << (x) << " "
#define PB push_back
#define SORT(obj) sort((obj).begin(), (obj).end())
#define SORTR(obj) sort((obj).begin(), (obj).end(), greater<>())
#define ALL(obj) (obj).begin(), (obj).end()
#define MOD 1000000007
#define PI (acos(-1))
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%MOD+MOD)%MOD){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= MOD) x -= MOD;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += MOD-a.x) >= MOD) x -= MOD;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= MOD; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime MOD
  mint inv() const { return pow(MOD-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
int in() {int x; cin>>x; return x;}
string stin() {string s; cin>>s; return s;}
/***** MAIN *****/
signed main() {
  int n = in();
  int k = in();
  int psts[n];
  REP(i,n) {
    cin >> psts[i];
  }
  int ans = 1000000000;
  REP(i,n-k+1) {
    int left = psts[i];
    int right = psts[i+k-1];
    int tmp_ans = abs(left) + abs(right - left);
    tmp_ans = min(tmp_ans, abs(right) + abs(right - left));
    ans = min(ans, tmp_ans);
  }
  cout << ans;
  cout << "\n";
  return 0;
}
/***** MAIN *****/