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 Q1 
{ 
	public static void main(String args[]) throws Exception 
	{
		Scan scan = new Scan();
		Print print = new Print();
		int N=scan.scanInt();
		int arr[] = new int[N];
		for (int i = 0; i < N; i++) 
		{
			arr[i] = scan.scanInt();
		}
		long ans=Integer.MAX_VALUE;
		long temp=0;
		for(int i=0;i<N;i++)
		{
			temp=0;
			for(int j=0;j<N;j++)
			{
				temp+=arr[j]*(Math.abs(i-j)+j+i)*2;
				//print.println(arr[j]*(Math.abs(i-j)+j+i)+" N "+i);	
			}
			ans=Math.min(ans,temp);
		}
		print.println(ans);
		print.close();
	}

	static class Print {
		private final BufferedWriter bw;

		public Print() {
			this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
		}

		public void print(Object object) throws IOException {
			bw.append("" + object);
		}

		public void println(Object object) throws IOException {
			print(object);
			bw.append("\n");
		}

		public void close() throws IOException {
			bw.close();
		}
	}

	static class Scan {
		private byte[] buf = new byte[1024 * 1024 * 4];
		private int index;
		private InputStream in;
		private int total;

		public Scan() {
			in = System.in;
		}

		public int scan() throws IOException {
			if (total < 0)
				throw new InputMismatchException();
			if (index >= total) {
				index = 0;
				total = in.read(buf);
				if (total <= 0)
					return -1;
			}
			return buf[index++];
		}

		public int scanInt() throws IOException {
			int integer = 0;
			int n = scan();
			while (isWhiteSpace(n))
				n = scan();
			int neg = 1;
			if (n == '-') {
				neg = -1;
				n = scan();
			}
			while (!isWhiteSpace(n)) {
				if (n >= '0' && n <= '9') {
					integer *= 10;
					integer += n - '0';
					n = scan();
				} else
					throw new InputMismatchException();
			}
			return neg * integer;
		}

		public double scanDouble() throws IOException {
			double doub = 0;
			int n = scan();
			while (isWhiteSpace(n))
				n = scan();
			int neg = 1;
			if (n == '-') {
				neg = -1;
				n = scan();
			}
			while (!isWhiteSpace(n) && n != '.') {
				if (n >= '0' && n <= '9') {
					doub *= 10;
					doub += n - '0';
					n = scan();
				} else
					throw new InputMismatchException();
			}
			if (n == '.') {
				n = scan();
				double temp = 1;
				while (!isWhiteSpace(n)) {
					if (n >= '0' && n <= '9') {
						temp /= 10;
						doub += (n - '0') * temp;
						n = scan();
					} else
						throw new InputMismatchException();
				}
			}
			return doub * neg;
		}

		public long scanLong() throws IOException {
			long ret = 0;
			long c = scan();
			while (c <= ' ') {
				c = scan();
			}

			boolean neg = (c == '-');
			if (neg) {
				c = scan();
			}

			do {
				ret = ret * 10 + c - '0';
			} while ((c = scan()) >= '0' && c <= '9');

			if (neg) {
				return -ret;
			}
			return ret;
		}

		public String scanString() throws IOException {
			StringBuilder sb = new StringBuilder();
			int n = scan();
			while (isWhiteSpace(n))
				n = scan();
			while (!isWhiteSpace(n) || n == ' ') {
				sb.append((char) n);
				n = scan();
			}
			return sb.toString();
		}

		private boolean isWhiteSpace(int n) {
			if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1)
				return true;
			return false;
		}
	}
}
