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.HashSet;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;

public class Main {


  private void solve(InputReader in, PrintWriter out) {
   int n = in.nextInt();
   int[] arr = new int[n];
    for (int i = 0; i < n; i++) {
      arr[i] = in.nextInt();
    }
    int res = Integer.MAX_VALUE;
    int count = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
          count += ((Math.abs(i - j) + Math.abs(j - 1) + Math.abs(1 - i) ) * 2) * arr[j - 1];
      }
      res = Math.min(res, count);
      count = 0;
    }
    out.print(res);
  }


  public static void main(String[] args) {
    new Main().run();
  }

  private void run() {

    try (InputReader in = new InputReader(System.in);
         PrintWriter out = new PrintWriter(System.out)) {
      solve(in, out);
    }
  }


  static class InputReader implements AutoCloseable {
    private BufferedReader reader;
    private StringTokenizer tokenizer;

    InputReader(InputStream stream) {
      //reader = new BufferedReader(new FileReader(stream), 32768);
      reader = new BufferedReader(new InputStreamReader(stream), 32768);
      tokenizer = null;
    }

    String next() {
      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
        try {
          tokenizer = new StringTokenizer(reader.readLine());
        } catch (IOException e) {
          throw new RuntimeException(e);
        }
      }
      return tokenizer.nextToken();
    }

    public int nextInt() {
      return Integer.parseInt(next());
    }

    public long nextLong() {
      return Long.parseLong(next());
    }

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

    @Override
    public void close() {
    }
  }

}
