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.math.BigInteger;
import java.io.*;

public class Codeforces{
	
	static class MyScanner{
		BufferedReader br;
		StringTokenizer st;
		
		MyScanner(){
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		MyScanner(FileReader fileReader){
			br = new BufferedReader(fileReader);
		}
		
		String nn(){
			while(st == null || !st.hasMoreElements()){
				try{
					st = new StringTokenizer(br.readLine());
				}catch(IOException e){
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		
		int ni(){
			return Integer.parseInt(nn());
		}
		
		double nd(){
			return Double.parseDouble(nn());
		}
		
		long nl(){
			return Long.parseLong(nn());
		}
	}
	
	private static PrintWriter out;
	
    public static void main(String[] args) throws IOException{
		
		// FileReader fileReader = new FileReader(new File("JavaFile.txt"));
		// out = new PrintWriter(new PrintStream("JavaOutputFile.txt"));
		// MyScanner sc = new MyScanner(fileReader);
		
		MyScanner sc = new MyScanner();
		out = new PrintWriter(new BufferedOutputStream(System.out));
		
		int n = sc.ni();
		int[] ar = new int[n + 1];
		for(int i = 1; i <= n; i++) ar[i] = sc.ni();
		int ans = Integer.MAX_VALUE;
		for(int i = 1; i <= n; i++){
			ans = Math.min(ans, getAns(n, ar, i));
		}
		
		out.println(ans);
		
		out.close();
    }
	
	private static int getAns(int n, int[] ar, int x){
		int ans = 0;
		for(int i = 1; i <= n; i++){
			ans += ar[i] * 2 * (Math.abs(i - x) + (i - 1) + (x - 1));
		}
		return ans;
	}
}