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.*;

public class A 
{
	
	
	static StringBuilder st = new StringBuilder();
	
	static int totalDist(int x , int i, int a)
	{
		
		return 2 * a *(Math.abs(x - i) + i + x ); 
	
	}
	
	public static void main(String[] args) throws Exception 
	{
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int n = sc.nextInt();
		int []a = new int [n];
		
		for(int i = 0 ; i < n ;i++) a[i] = sc.nextInt() ; 
		
		int ans = Integer.MAX_VALUE ; 
		
		for(int x = 0 ; x < n ; x++)
		{	
			int sum = 0 ;
		
			for(int i = 0 ; i < n ;i++)
				sum += totalDist(x, i ,a[i]);
			
			ans = Math.min(ans, sum);
		}
		
		out.println(ans);
		out.flush();
		out.close();
		
		
	}
	
	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(String path) throws Exception {
			br = new BufferedReader(new FileReader(path));
		}

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

	}

	private static  boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	private static  void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); }

}