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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
const int N = 5005;
const long long mod = 100000007;
int n, x, y;
int size[N];
long long tmp[N], df[N];
long long f[N][N];
std::vector<int> to[N];
inline void dfs(int u, int fa) {
    f[u][1] = 1;
    size[u] = 1;
    for (auto v: to[u]) {
        if (v == fa) continue;
        dfs(v, u);
        for (int i = 0; i <= size[u]; ++i) tmp[i] = f[u][i], f[u][i] = 0;
        for (int i = 0; i <= size[u]; ++i) {
            for (int j = 0; j <= size[v]; ++j) {
                f[u][i + j] += tmp[i] * f[v][j] % mod;
            }
        }
        size[u] += size[v];
        for (int i = 0; i <= size[u]; ++i) f[u][i] %= mod;
    }
    for (int i = 2; i <= size[u]; i += 2) f[u][0] += -df[i - 1] * f[u][i] % mod;
    f[u][0] %= mod;
}
int main() {
    std::cin >> n;
    for (int i = 1; i < n; ++i) {
        std::cin >> x >> y;
        to[x].push_back(y);
        to[y].push_back(x);
    }
    df[0] = df[1] = 1;
    for (int i = 2; i <= n; ++i) df[i] = df[i - 2] * i % mod;
    dfs(1, 1);
    std::cout << -f[1][0] << std::endl;
    return 0;
}