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.io.*;
import java.math.*;
 
public class A
{
    public static void process(int test_number)throws IOException
    {
        int n = ni(), arr[] = new  int[n+1];
        for(int i = 1; i <= n; i++)
            arr[i] = ni();

        int mn = oo, total = 0;

        for(int x = 1; x <= 101; x++){
            total = 0;

            for(int j = 1; j <= n; j++){
                total += arr[j] * (Math.abs(x - j) + (j - 1)  + (x - 1) + (j - 1) +
                                    Math.abs(x - j) + (x - 1));
            }

            mn = Math.min(mn, total);
        } 
        pn(mn);
    }
    
    static final int oo = Integer.MAX_VALUE;
    static final long mod = (long)1e9+7l;
    
    static FastReader sc;
    static PrintWriter out;
    public static void main(String[]args)throws IOException
    {
        out = new PrintWriter(System.out);
        sc = new FastReader();
 
        long s = System.currentTimeMillis();
        int t = 1;
       // t = ni();
        for(int i = 1; i <= t; i++)
            process(i);
 
        out.flush();
        System.err.println(System.currentTimeMillis()-s+"ms");
    }

    static void trace(Object... o){ System.err.println(Arrays.deepToString(o)); };
    static void pn(Object o){ out.println(o); }
    static void p(Object o){ out.print(o); }
    static int ni()throws IOException{ return Integer.parseInt(sc.next()); }
    static long nl()throws IOException{ return Long.parseLong(sc.next()); }
    static double nd()throws IOException{ return Double.parseDouble(sc.next()); }
    static String nln()throws IOException{ return sc.nextLine(); }
    static long gcd(long a, long b)throws IOException{ return (b==0)?a:gcd(b,a%b);}
    static int gcd(int a, int b)throws IOException{ return (b==0)?a:gcd(b,a%b); }
    static int bit(long n)throws IOException{ return (n==0)?0:(1+bit(n&(n-1))); }
    
    static class FastReader{ 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader(){ 
            br = new BufferedReader(new InputStreamReader(System.in)); 
        } 
  
        String next(){ 
            while (st == null || !st.hasMoreElements()){ 
                try{ st = new StringTokenizer(br.readLine()); } catch (IOException  e){ e.printStackTrace(); } 
            } 
            return st.nextToken(); 
        } 
  
        String nextLine(){ 
            String str = ""; 
            try{ str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } 
            return str; 
        } 
    } 
}
