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.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.*;

public class TheFairNutAndElevator {

	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner() {
			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();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	static class Pair {
		int diff;
		int count;

		public Pair(int diff, int count) {
			super();
			this.diff = diff;
			this.count = count;
		}

	}

	public static void main(String[] args) {

		FastScanner f = new FastScanner();
		PrintWriter p = new PrintWriter(System.out);
		int n = f.nextInt();
		long arr[] = new long[n + 1];
		for (int i = 1; i <= n; i++) {
			arr[i] = f.nextLong();
		}
		long max = Long.MIN_VALUE;
		long maxind = -1;
		for (int i = 2; i <= n; i++) {
			if (arr[i] > max) {
				max = arr[i];
				maxind = i;
			}
		}
		long x = maxind;
		long ans = 0;
		for (int i = 2; i <= n; i++) {
			ans += (Math.abs(x - (long) i) + Math.abs(1 - (long) i) + Math.abs(1 - x)) * 2 * arr[i];
		}

		if (x == -1)
			p.println(0);
		else
			p.println(ans);
		p.close();

	}

	/// Ceil
	static long ceil(long x, long m) {
		long res = x / m;
		if (x % m != 0) {
			res++;
		}
		return res;
	}

	// ------------------------------------------------------------------------------------------------
	// makes the prefix sum array
	static long[] prefixSum(long arr[], int n) {
		long psum[] = new long[n];
		psum[0] = arr[0];
		for (int i = 1; i < n; i++) {
			psum[i] = psum[i - 1] + arr[i];
		}
		return psum;
	}
	// ------------------------------------------------------------------------------------------------

	// makes the suffix sum array
	static long[] suffixSum(long arr[], int n) {
		long ssum[] = new long[n];
		ssum[n - 1] = arr[n - 1];
		for (int i = n - 2; i >= 0; i--) {
			ssum[i] = ssum[i + 1] + arr[i];
		}
		return ssum;
	}

//------------------------------------------------------------------------------------------
	// BINARY EXPONENTIATION OF A NUMBER MODULO M FASTER METHOD WITHOUT RECURSIVE
	// OVERHEADS
	static long m = (long) (1e9 + 7);

	static long binPower(long a, long n, long m) {
		if (n == 0)
			return 1;
		long res = 1;
		while (n > 0) {
			if ((n & 1) != 0) {
				res *= a;
			}
			a *= a;
			n >>= 1;
		}
		return res;
	}

//-------------------------------------------------------------------------------------------

	// gcd
	static long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}

//------------------------------------------------------------------------------------------

	// lcm
	static long lcm(long a, long b) {
		return a / gcd(a, b) * b;
	}

//------------------------------------------------------------------------------------------
	// BRIAN KERNINGHAM TO CHECK NUMBER OF SET BITS
	// O(LOGn)
	static int setBits(int n) {
		int count = 0;
		while (n > 0) {
			n = n & (n - 1);
			count++;
		}
		return count;
	}
//------------------------------------------------------------------------------------------

//------------------------------------------------------------------------------------------
	// 0 based indexing
	static boolean KthBitSet(int n, int k) {
		int mask = 1;
		mask = mask <<= k;
		if ((mask & n) != 0)
			return true;
		else
			return false;
	}

//------------------------------------------------------------------------------------------

	// EXTENDED EUCLIDEAN THEOREM
	// TO REPRESENT GCD IN TERMS OF A AND B
	// gcd(a,b) = a.x + b.y where x and y are integers

	static long x = -1;
	static long y = -1;

	static long gcdxy(long a, long b) {
		if (b == 0) {
			x = 1;
			y = 0;
			return a;
		} else {
			long d = gcdxy(b, a % b);
			long x1 = y;
			long y1 = x - (a / b) * y;
			x = x1;
			y = y1;

			return d;
		}
	}

//-------------------------------------------------------------------------------------------------F
}
