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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class _template {
    static BufferedReader _in = new BufferedReader(new InputStreamReader(System.in));
    static StringTokenizer _stk;
    static String next() {
        try {
            while (_stk == null || !_stk.hasMoreTokens())
                _stk = new StringTokenizer(_in.readLine());

            return _stk.nextToken();
        } catch(Exception ex) {
            throw new RuntimeException("Failed reading input", ex);
        }
    }
    static int nextInt() { return Integer.parseInt(next()); }
    static long nextLong() { return Long.parseLong(next()); }
    static double nextDouble() { return Double.parseDouble(next()); }

    public static void main(String [] args) throws IOException {
        int N = nextInt();
        int A[] = new int[N];
        for(int i=0; i<N; i++)
            A[i] = nextInt();

        int xbest = -1;
        int best = (int)1e9;
        for(int x=1; x<=N; x++) {
            int v = calc(A, x);
            if(v<best) {
                best = v;
                xbest = x;
            }
        }

        System.out.println(best);
    }

    static int calc(int A[], int X) {
        int sum = 0;
        for(int i=0; i<A.length; i++) {
            int xtofloor = Math.abs(i+1 - X);
            int floortoground = Math.abs(i+1 - 1);
            int groundtox = X-1;

            int xtoground = X-1;
            int groundtofloor = i;
            int floortox = xtofloor;
            int total = xtofloor+floortoground+groundtox;
            sum += total*2*A[i];
        }
        return sum;
    }


}