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 5.4.1) (judge verdict: Wrong Answer)

#include<cstdio>

using namespace std;

int abs(int x){
    if (x > 0) {
        return x;
    }else{
        return -x;
    }
    
}

int main(){
    int n, k;
    scanf("%d %d", &n, &k);
    int x[n];
    for(int i = 0; i < n; i++){
        scanf("%d", &x[i]);
    }
    int idx0;
    for(int i = 0; i < n; i++){
        if (x[i] >= 0) {
            idx0 = i;
            break;
        }
    }
    
    int mn = 500000000;
    for(int i = 0; i < k; i++){
        if (idx0 + i >= n ) {
            break;
        }else if(idx0 + i + 1 < k){
            continue;
        }
        int sm = 0;
        sm = 2 * x[idx0 + i] + abs(x[idx0 + i + 1 - k]);
        if (mn > sm) {
            mn = sm;
        }
    }
    // printf("%d\n", mn);
    if (n - idx0 >= k - 1) {
        int sm = x[idx0 + k - 1];
        if (mn > sm) {
            mn = sm;
        }   
    }

    if (x[idx0] != 0) {
        idx0--;
    }
    
    // printf("%d\n", mn);
    // printf("idx0 = %d\n", idx0);
    for(int i = 0; i < k; i++){
        if (idx0 - i < 0) {
            break;
        }else if(n - idx0 + i + 1 < k){
            continue;
        }
        int sm = 0;
        sm = 2 * abs(x[idx0 - i]) + x[idx0 + k - i - 1];
        if (mn > sm) {
            mn = sm;
        }
        // printf("sm = %d\n", sm);
    }
    // printf("%d\n", mn);
    // printf("%d\n", idx0);
    if (idx0 >= k - 1) {
        int sm = abs(x[idx0 - k + 1]);
        if (mn > sm) {
            mn = sm;
        }   
    }
    printf("%d\n", mn);
}