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.
/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */


import java.util.Arrays;
import java.util.Scanner;

/**
 *
 * @author Amr Elkady
 */
public class CodeForces {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
        
        Scanner sc= new Scanner(System.in);
        
        int n=sc.nextInt();
        
        int a[];
        a=new int [n];
        
        int maxx=0;
        int xfloor=-1;
        for(int i =0 ;i<n;i++)
        {
            a[i]=sc.nextInt();
            
            if(a[i]>maxx)
            {
                maxx=a[i];
                xfloor=i+1;
            }
                
        }
        System.out.println(maxx+"<<<>>>"+xfloor);
        
        int mincost=0;
        for(int i=1;i<=a.length;i++)
        {
            int cost;
            cost=Math.abs((i)-xfloor)+Math.abs(i-1)+Math.abs(1-xfloor);
             System.out.println(cost*a[i-1]);

            
             
                mincost+=cost*a[i-1]*2;
           
        }
         
         
        System.out.println(mincost);
        
        
       
        
        
    }
    
    static int maxMin(int k, int[] arr) {
        
       
        int min=Integer.MAX_VALUE;
         for(int i=0; i<=arr.length-k;i++)
         {
             int temp=arr[i+k-1]-arr[i];
            // System.out.println(arr[i]+"<<<>>>"+arr[i+k-1]);
             if (min>temp){
                 min=temp;
                // System.out.println("here");
             }
             
             temp=0;
         }        
        
        return min;

    }
    
    
}
