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 sys


def solve(io):
	N = io.read_int()
	A = io.read_int_array(N)

	best = 1234567890
	for x in range(0, N):
		score = calculate(x, A)
		best = min(best, score)
	io.println(best)


def calculate(x, A):
	cnt = 0
	for f in range(0, len(A)):
		cnt += 2 * A[f] * (abs(x - f) + f + x)
	return cnt


# +---------------------+
# | TEMPLATE CODE BELOW |
# |    DO NOT MODIFY    |
# +---------------------+


class IO:
	in_stream = None
	out_stream = None
	raw = ""
	buf = []
	pos = 0

	def __init__(self, input_stream, output_stream):
		self.in_stream = input_stream
		self.out_stream = output_stream

	def read_to_buffer(self):
		self.raw = self.in_stream.readline().rstrip('\n')
		self.buf = self.raw.split()
		self.pos = 0

	def read_string(self):
		while self.pos == len(self.buf):
			self.read_to_buffer()
		ans = self.buf[self.pos]
		self.pos += 1
		return ans

	def read_int(self):
		return int(self.read_string())

	def read_float(self):
		return float(self.read_string())

	def read_string_array(self, N, offset=0):
		arr = [None] * offset
		for _ in range(0, N):
			arr.append(self.read_string())
		return arr

	def read_int_array(self, N, offset=0):
		arr = [None] * offset
		for _ in range(0, N):
			arr.append(self.read_int())
		return arr

	def read_float_array(self, N, offset=0):
		arr = [None] * offset
		for _ in range(0, N):
			arr.append(self.read_float())
		return arr

	def read_line(self):
		while self.pos == len(self.buf):
			self.read_to_buffer()
		if self.pos > 0:
			raise ValueError("Cannot call read_line in the middle of a line.")
		self.pos = len(self.buf)
		return self.raw

	def print(self, s):
		self.out_stream.write(str(s))

	def println(self, s):
		self.print(s)
		self.print('\n')
	
	def println_array(self, arr, sep = ' '):
		self.println(sep.join(str(x) for x in arr))

	def flush_output(self):
		self.out_stream.flush()


pythonIO = IO(sys.stdin, sys.stdout)
solve(pythonIO)
pythonIO.flush_output()
