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 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <ctime>
#include <cstring>
#include <functional>
#include <iostream>
#include <iomanip>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <regex>
#include <vector>

#define fix(n)          cout<<fixed<<setprecision(n)
#define rep(i,n)        for (int i = 0; i < (n); ++i)
#define all(a)          (a).begin(), (a).end()
#define sort(a)         sort(all(a))
#define uniq(a)         sort(a);(a).erase(unique(all(a), (a).end())
#define reverse(a)      reverse(all(a))
#define ctos(c)         string(1, (c))
#define out(d)          cout << (d)
#define outl(d)         std::cout<<(d)<<"\n"
#define YES()           cout << "YES" << endl
#define NO()            cout << "NO" << endl
#define Yes()           cout << "Yes" << endl
#define No()            cout << "No" << endl
#define ceil(x, y)      ((x + y - 1) / (y))
#define debug(x)        cerr << #x << ": " << (x) << '\n'
#define debug2(x, y)    cerr << #x << ": " << (x) << ", " << #y << ": " << (y) << '\n'
#define debug3(x, y, z) cerr << #x << ": " << (x) << ", " << #y << ": " << (y) << ", " << #z << ": " << (z) << '\n'
#define dbg(v)          for (size_t _ = 0; _ < v.size(); ++_){ cerr << #v << "[" << _ << "] : " << v[_] << '\n'; }
#define pb              push_back
#define fst             first
#define int             long long

using namespace std;
using ll = long long;
using ld = long double;
using P = pair<ll,ll>;

const ll MOD = 1000000007; // 10^9 + 7
const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

void solve() {
  ll n, k; cin >> n >> k;
  vector<ll> x(n); rep(i, n) cin >> x[i];
  int t = distance(x.begin(), lower_bound(all(x), 0));

  ll ans = 0xffffffff;
  for (int a = 1; a < k; a++) {
    int b = k - a;
    int li = t-(a-1);
    int ri = t+b;
    if (li >= 0 && ri < n) {
      ans = min({
        ans,
        2*abs(x[li]) + abs(x[ri]),
        abs(x[li]) + 2*abs(x[ri])
      });
    }
  }
  if (t+k-1 < n) ans = min(ans, abs(x[t+k-1]));
  if (t-k >= 0)  ans = min(ans, abs(x[t-k]));
  outl(ans);
}

signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  srand((unsigned)time(NULL));
  fix(12);

  solve();
}