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 ;
import java.io.*;
import java.util.*;

public class Problem_1084A
{
	public static void main(String[] args) throws IOException 
	{
		Scanner sc = new Scanner();
		PrintWriter pw = new PrintWriter(System.out);
		int n=sc.nextInt();
		int[]a=new int[n+1];
		for(int i=1;i<=n;i++)a[i]=sc.nextInt();
		long ans=Long.MAX_VALUE;
		for(int i=1;i<=n;i++)
		{
			long tmp=0;
			for(int j=1;j<=n;j++)
			{
				tmp+=Math.abs(j-i)*a[j] + (j-1)*a[j] +(i-1)*a[j];
				tmp+=(i-1)*a[j] + (j-1)*a[j] + Math.abs(j-i)*a[j] ;
//				System.out.println(i+" "+a[j]*j+" "+Math.abs(j-i)*a[j]+" "+i*a[j]);
			}
			ans=Math.min(ans, tmp);
		}
		pw.print(ans);
		pw.close();    
	}
	static class Scanner {
		BufferedReader br;
		StringTokenizer st;
		
		Scanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		String next() throws IOException {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(br.readLine());
			}
			return st.nextToken();
		}
		
		int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
		
		long nextLong() throws IOException {
			return Long.parseLong(next());
		}
		
		double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
		
		String nextLine() throws IOException {
			return br.readLine();
		}
		boolean hasnext() throws IOException{
			return br.ready();
		}
		
	}
}
