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.util.*;
import java.awt.geom.*;
import java.io.*;
import java.math.*;

public class Main
{

  public static void main(String[] args) throws Exception
  {
    long startTime = System.nanoTime();
    int n=in.nextInt();
    int [] people = new int[n];
    for(int i=0;i<n;i++)
        people[i]=in.nextInt();
    int ans = Integer.MAX_VALUE;
    for(int i=1;i<=n;i++){
        ans = Math.min(ans,drive(people,i));
    }
    out.println(ans);
    long endTime = System.nanoTime();   
    err.println("Execution Time : +" + (endTime-startTime)/1000000 + " ms");
    exit(0);
  }

  static int drive(int [] people,int x){
     int ans = 0;
     for(int i=0;i<people.length;i++){
        int a = (i+1);
        int b = 1;
        int c = 1;
        int d = (i+1);
        ans += energy(a,b,x)*people[i];
        ans += energy(c,d,x)*people[i];
     }
     return ans;
  }

  static int energy(int a,int b,int x){
     return Math.abs(a-x) + Math.abs(a-b) + Math.abs(b-x);
  }

  static class InputReader
  {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

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

    static void exit(int a)
    {
      out.close();
      err.close();
      System.exit(a);
    }

    static InputStream inputStream = System.in;
    static OutputStream outputStream = System.out;
    static OutputStream errStream = System.err;
    static InputReader in = new InputReader(inputStream);
    static PrintWriter out = new PrintWriter(outputStream);
    static PrintWriter err = new PrintWriter(errStream);
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

}
