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.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
import java.util.Scanner;
import java.util.Vector;

/**
 * necklace -
 * pearls o
 * @author Rodrigo
 *
 */
public class Main {
	static Scanner sc = new Scanner(System.in);

	private static void multipleEntrace(){
		while(sc.hasNextLine()){

			Integer numeroAndare = sc.nextInt();
			String linha = sc.nextLine();
			linha = sc.nextLine();
			String[] pessoas = linha.split(" ");
			List<Integer> lista = new Vector<>();
			for(String pAndar:pessoas) {
				lista.add(Integer.parseInt(pAndar));
			}
			
			if(numeroAndare==1) {
				System.out.println(pessoas.length*2);
			} else {
				Integer between = numeroAndare+1;
				Integer multipleAndar = 0;
				Integer sum =0;
				for(int i=0 ;i<numeroAndare.intValue();i++) {
					Integer pAndar = lista.get(i).intValue();
					if(i==1) {
						multipleAndar=2;
					}
					if(i==0) {
						sum+= ((pAndar*numeroAndare));
					} else {
						
						sum+= ((pAndar.intValue()*multipleAndar.intValue()*2));		
						multipleAndar=multipleAndar*2;

					}
					
				}
				System.out.println(sum);
			}
			
		}
	}
	private static void OneByOneEntrace(){

	}
	
	public static void main(String[] args) {
		multipleEntrace();
//		OneByOneEntrace();
	}
	
	public static Integer nPositivo(Integer x){
		return Math.abs(x);
	} 
	
	public static String reverse (String x){
		return new StringBuilder(x).reverse().toString();
	}
	public static List<Long> reverse (List<Long> x){
		Collections.reverse(x);
		return x;
	 }
	public static void p(StringBuilder x){
		p(x.toString());
	}
	public static void p(String x){
		System.out.println(x);
	}
	public static void p(Long x){
		System.out.println(x);
	}	
	public static void p(Integer x){
		System.out.println(x);
	}
	public static void p(List<Integer> x){
		System.out.println(x);
	}
	public static void pNoBorder(List<Integer> x){
		System.out.println(x.toString().substring(1, x.toString().length()-1));
	}
}

