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.*;

public class Try {

	public static void main(String[] args) throws IOException {
		Scanner sc=new Scanner(System.in);
		PrintWriter out =new PrintWriter (System.out);
		int n=sc.nextInt();
		int []a=new int[n];
		for(int i=0;i<n;i++)
			a[i]=sc.nextInt();
		long min=(long)Double.POSITIVE_INFINITY;
        for(int i=0;i<n;i++) {
        	int y=i; long tr=0;
        	for(int j=0;j<n;j++) {
        		int l=0;
        		int p=1;
        		l=Math.abs(i-j)+j+i;	
        		//System.out.println(l);
        		p=a[j]*l*2;
        		tr+=p;
        		//System.out.println(tr+"j");
        	}
        	if(tr<min)
        		min=tr;
        }
        out.println(min);
        out.flush();
	}
	static	class Scanner{
        StringTokenizer st;
        BufferedReader br;
        public Scanner(InputStream system) {br = new BufferedReader(new InputStreamReader(system));}
        public Scanner(String file) throws Exception {br = new BufferedReader(new FileReader(file));}
        public String next() throws IOException
        {
            while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }
        public String nextLine()throws IOException{return br.readLine();}
        public int nextInt() throws IOException {return Integer.parseInt(next());}
        public double nextDouble() throws IOException {return Double.parseDouble(next());}
        public char nextChar()throws IOException{return next().charAt(0);}
        public Long nextLong()throws IOException{return Long.parseLong(next());}
        public boolean ready() throws IOException{return br.ready();}
        public void waitForInput() throws InterruptedException {Thread.sleep(3000);}
    }
}
