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 org.omg.CORBA.MARSHAL;

import java.io.*;
import java.util.*;

public class Banhi {
    public  static  void  main(String[]args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int n = sc.nextInt();
        int [] ev = new int[n];
        for (int i =0;i<n;i++)ev[i]=sc.nextInt();
        int sum =Integer.MAX_VALUE;
        for (int i =1;i<=100;i++){
            int cur=0;
            for (int j =1;j<n;j++){
                cur+=(Math.abs(i-j)+j+i)*ev[j];
            }
            cur*=2;
            //System.out.println(cur+" "+i);
            sum=Math.min(sum,cur);
        }

        pw.println(sum);
        pw.flush();
    }
    static class Scanner
    {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

        public String next() throws IOException
        {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public int nextInt() throws IOException {return Integer.parseInt(next());}

        public long nextLong() throws IOException {return Long.parseLong(next());}

        public String nextLine() throws IOException {return br.readLine();}

        public double nextDouble() throws IOException
        {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if(x.charAt(0) == '-')
            {
                neg = true;
                start++;
            }
            for(int i = start; i < x.length(); i++)
                if(x.charAt(i) == '.')
                {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                }
                else
                {
                    sb.append(x.charAt(i));
                    if(dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg?-1:1);
        }

        public boolean ready() throws IOException {return br.ready();}


    }
}