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.
/* package whatever; // don't place package name! */
 
import java.util.*;
import java.io.*;
 
/* Name of the class has to be "Main" only if the class is public. */
public class yay
{
	public static int gcd(int a, int b)
	{
		if(b==0) return a;
		return gcd(b, a%b);
	}
	public static int lcm(int a, int b)
	{
	    return a * (b / gcd(a, b));
	}
	public static void main(String[] args) throws java.lang.Exception{
		MyScanner sc = new MyScanner();
		PrintWriter out = new PrintWriter(System.out);
		long n = sc.nextLong();
		long[] arr = new long[(int)n+1];
		long max = 0;
		long index = 0;
		for(int i = 1;i<=n;i++)
		{
			arr[i] = sc.nextLong();
			if(max<arr[i])
			{
				max = arr[i];
				index = i;
			}
		}
		long sum = 0;
		for(int i = 1;i<=n;i++)
		{
			if(i>index)
				sum+=(((index-1)*4)+(i-index)*4)*arr[i];
			else
				sum+=((index-1)*4)*arr[i];
		}
		out.println(sum);
		out.close();
    }

	static class MyScanner{
        StringTokenizer st;
        BufferedReader br;
        public MyScanner(){br = new BufferedReader(new InputStreamReader(System.in));}
        public MyScanner(FileReader s) throws FileNotFoundException {	br = new BufferedReader(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 { return Double.parseDouble(next()); }
        public boolean ready() throws IOException {return br.ready();}
    }
} 