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.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.ByteArrayInputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * @author tainic on Nov 28, 2018
 */
public class A {

    //private static final boolean LOCAL = "aurel".equalsIgnoreCase(System.getenv().get("USER"));
    private static final boolean LOCAL = false;

    private static final String TEST =
        "2\n" +
        "1 1";

    void solve(Scanner in, PrintWriter out) {
        int n = in.nextInt();
        int[] a = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            a[i] = in.nextInt();
        }

        int min = Integer.MAX_VALUE;
        for (int x = 1; x <= n; x++) {
            min = Math.min(f(a, x), min);
        }
        out.println(min);
    }

    int f(int[] a, int x) {
        int sum = 0;
        for (int i = 1; i < a.length ; i++) {
            sum += 2 * a[i] * (Math.abs(x - i) + Math.abs(i - 1) + Math.abs(x - 1));
        }
        return sum;
    }

    //region main
    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedInputStream(!LOCAL ? System.in : new ByteArrayInputStream(TEST.getBytes())));
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out, 2048), false);
        new A().solve(in, out);
        out.close();
    }
    //endregion

    //region imports
    //endregion

}