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.io.*;

public class Main{



   public static long primeFactors(long n) 
{ 
    
    // Print the number of 2s that divide n 
  
    while (n%2 == 0) 
    { 
        return 2;
      
    } 
  
    // n must be odd at this point.  So we can skip  
    // one element (Note i = i +2) 
    for (long i = 3; i <= Math.sqrt(n); i = i+2) 
    { 
        // While i divides n, print i and divide n 
        while (n%i == 0) 
        { 
            return i;
            
             
        } 
    } 
  
    // This condition is to handle the case when n  
    // is a prime number greater than 2 
    if (n > 2) 
        return n;

    return 0;
       
} 

    

    public static boolean isPrime(long n) 
    { 
        // Corner cases 
        if (n <= 1) return false; 
        if (n <= 3) return true; 
      
        // This is checked so that we can skip  
        // middle five numbers in below loop 
        if (n % 2 == 0 || n % 3 == 0) return false; 
      
        for (long i = 5; i * i <= n; i = i + 6) 
            if (n % i == 0 || n % (i + 2) == 0) 
            return false; 
      
        return true; 
    } 

    public static void main(String[] args) {

        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader inp = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Solver solver = new Solver();
        solver.solve(inp, out);
        out.close();
        
          }



    static class Solver {
        private void solve(InputReader inp, PrintWriter out1) {

        int n = inp.nextInt();
        int arr[] = new int[n];

        for(int i=0;i<n;i++)
        {
            arr[i] = inp.nextInt();
        }
        int min=Integer.MAX_VALUE;
        for(int i=0;i<n;i++) //ith floor lift
        {
            int x=0;
            for(int j=0;j<n;j++)
            {
                x+=2*arr[j]*(Math.abs(j-i)+i+j);
               
            }
            // out1.println(x+" x");
            if(x<min)
                min=x;
        }
        out1.println(min);
    }



    
}






    static class InputReader {
        BufferedReader reader;
        StringTokenizer tokenizer;

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

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

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

        public long nextLong() {
            return Long.parseLong(next());
        }
    }
}
class ele{
    long value;
    long i;
    boolean flag;
    public ele(long value,long i)
    {
        this.value = value;
        this.i=i;
        this.flag = false;
    }
}