The Fair Nut lives in n story house. a_i people live on the i-th floor of the house. Every person uses elevator twice a day: to get from the floor where he/she lives to the ground (first) floor and to get from the first floor to the floor where he/she lives, when he/she comes back home in the evening. 

It was decided that elevator, when it is not used, will stay on the x-th floor, but x hasn't been chosen yet. When a person needs to get from floor a to floor b, elevator follows the simple algorithm: 

  * Moves from the x-th floor (initially it stays on the x-th floor) to the a-th and takes the passenger. 
  * Moves from the a-th floor to the b-th floor and lets out the passenger (if a equals b, elevator just opens and closes the doors, but still comes to the floor from the x-th floor). 
  * Moves from the b-th floor back to the x-th. 

The elevator never transposes more than one person and always goes back to the floor x before transposing a next passenger. The elevator spends one unit of electricity to move between neighboring floors. So moving from the a-th floor to the b-th floor requires |a - b| units of electricity.

Your task is to help Nut to find the minimum number of electricity units, that it would be enough for one day, by choosing an optimal the x-th floor. Don't forget than elevator initially stays on the x-th floor. 

Input

The first line contains one integer n (1 ≤ n ≤ 100) — the number of floors.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 100) — the number of people on each floor.

Output

In a single line, print the answer to the problem — the minimum number of electricity units.

Examples

Input

3
0 2 1


Output

16

Input

2
1 1


Output

4

Note

In the first example, the answer can be achieved by choosing the second floor as the x-th floor. Each person from the second floor (there are two of them) would spend 4 units of electricity per day (2 to get down and 2 to get up), and one person from the third would spend 8 units of electricity per day (4 to get down and 4 to get up). 4 ⋅ 2 + 8 ⋅ 1 = 16.

In the second example, the answer can be achieved by choosing the first floor as the x-th floor.
import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

/**
 * Created by Mach on 10.12.2018.
 */
public class round526a {
    public static void main(String[] args) {
        Reader526a r=new Reader526a();
        r.init(System.in);

        int n=r.nextInt();
        int a[]=new int[n+1];
        for(int i=1; i<=n; i++) a[i]=r.nextInt();
        if (n==1) {
            System.out.println("0");
            return;
        }

        int min=-1;
        for(int x=1; x<=n; x++){
            int res=0;
            int dif;
            for(int j=1; j<=n; j++) {
                dif=(x>j)?(x-j):(j-x);
                res=res+(dif+(j-1)+(x-1))*2*a[j];
            }

            if (min<0) min=res;
            if (min>res) min=res;
        }
        System.out.println(min);
    }
}

class Reader526a {
    static BufferedReader reader;
    static StringTokenizer tokenizer;

    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
        reader = new BufferedReader(
                new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }

    /** get next word */
    static String next() /* throws IOException */ {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            try {
                tokenizer = new StringTokenizer(
                        reader.readLine());
            } catch(Exception e) {e.printStackTrace();}
        }
        return tokenizer.nextToken();
    }

    static int nextInt() {
        return Integer.parseInt( next() );
    }

    static long nextLong(){ return (long)nextInt();}

    static double nextDouble() {
        return Double.parseDouble( next() );
    }

}