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>
#define rep(i,n)for(int i=0;i<n;++i)
#include<string>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<int, int>P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
//max=({});
//条件式が真ならwhileの中身を回し続ける
//printf("%d\n", ans);

//pairの入力
//vector<pair<ll, ll>>work(n);
//rep(i, n) {
//	ll a, b;
//	cin >> a >> b;
//	work[i] = make_pair(a, b);
//for(auto p:mp)(mapの探索)
//printf("%.10f\n",なんちゃら)

int g[15][15];
const int INF = 1001001001;
const int dx[4] = { -1,0,1,0 };
const int dy[4] = { 0,-1,0,1 };

//最大公約数
ll gcd(ll x, ll y) {
	return y ? gcd(y, x % y) : x;
}

ll lcm(ll x, ll y) {
	if (x == 0 || y == 0)return 0;
	return (x / gcd(x, y) * y);
}

//素因数分解
vector<pair<ll, int>>factorize(ll n) {
	vector<pair<ll, int>>res;
	for (ll i = 2;i * i <= n;++i) {
		if (n % i)continue;
		res.emplace_back(i, 0);
		while (n % i == 0) {
			n /= i;
			res.back().second++;
		}
	}
	if (n != 1)res.emplace_back(n, 1);
	return res;
}

vector<ll>relation[20], money(20, 1);

int main() {
	int n, k;
	cin >> n >> k;
	vector<int>a(n), b, c;
	rep(i, n) {
		cin >> a[i];
	}
	int ans = 0;
	rep(i, k) {
		ans += abs(a[i]);
	}
	int res = ans;
	for (int i = 0;i < n - k;++i) {
		res += abs(a[k + i]);
		res -= abs(a[i]);
		ans = min(ans, res);
	}
	cout << ans << endl;
	return 0;
}