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.*;
import java.util.*;

/**
 * Built using my Brain
 * Actual solution is at the bottom
 *
 * @author Lenard Hoffstader
 */

public class cfjava
{
   public static void main(String[] args)
   {
      OutputStream outputStream = System.out;
      FastReader in = new FastReader();
      PrintWriter out = new PrintWriter(outputStream);
      PROBLEM solver = new PROBLEM();
      int t=1;
      // t=in.nextInt();
      for(int i=0;i<t;i++){
         solver.solve(in, out);
      }
      out.close();
   }

   static class PROBLEM
   {
      static int getUnits(int arr[],int x){
         int res=0;
         for(int i=1;i<=arr.length;i++){
            res+=2*(Math.abs(i-x)+i-1+Math.abs(x-1))*arr[i-1];
         }
         return res;
      }
      public void solve(FastReader in,PrintWriter out)
      {
         int n=in.nextInt();
         int arr[]=new int[n];
         for(int i=0;i<n;i++){arr[i]=in.nextInt();}
         int money=Integer.MAX_VALUE;
         for(int i=1;i<=n;i++){
            money=Math.min(money, getUnits(arr,i));
         }
         out.println(money);
      }
   }

   static class FastReader 
   { 
      BufferedReader br; 
      StringTokenizer st; 

      public FastReader() 
      { 
         br = new BufferedReader(new InputStreamReader(System.in)); 
      } 

      String next() 
      { 
         while (st == null || !st.hasMoreElements()) 
         {
            try
            { 
               st = new StringTokenizer(br.readLine()); 
            } 
            catch (IOException  e) 
            { 
               e.printStackTrace(); 
            } 
         } 
         return st.nextToken(); 
      } 

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

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

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

      boolean nextBoolean(){return Boolean.parseBoolean(next());}

      String nextLine() 
      { 
         String str = ""; 
         try
         { 
            str = br.readLine(); 
         } 
         catch (IOException e){e.printStackTrace();} 
         return str; 
      } 
   }
}