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.*;
public class lol {
	
	public static void main (String [] args)throws IOException,InterruptedException {
	     Scanner sc= new Scanner(System.in) ;
	     int n =sc.nextInt();
	     int [] a = new int [n];
	     for(int i = 0;i<n;i++) {
	    	 a[i]=sc.nextInt();
	     }
	     int comp = 20000;
	     int c = 0;
	     int x ;
	     for(int i = 0;i<n;i++) {
	    	 x=i;
	    	 for(int j = 0;j<n;j++) {
	    		 c+=(Math.abs(x-j)+Math.abs(x-0)+j)*a[j]*2;
	    		 
	    	 }
	    	 if(c<comp)
	    		 comp=c;
	    	 c=0;
	     }
	     System.out.println(comp);
	  } 	 
	     
	      	
	 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);
			
		}
		}
}