Problem: You have a pot and N ingredients. Each ingredient has a real number parameter called value, and the value of the i-th ingredient (1 \leq i \leq N) is v_i.

When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.

After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Constraints

* 2 \leq N \leq 50
* 1 \leq v_i \leq 1000
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
v_1 v_2 \ldots v_N


Output

Print a decimal number (or an integer) representing the maximum possible value of the last ingredient remaining.

Your output will be judged correct when its absolute or relative error from the judge's output is at most 10^{-5}.

Examples

Input

2
3 4


Output

3.5


Input

3
500 300 200


Output

375


Input

5
138 138 138 138 138


Output

138

Solution (in PYTHON):

Solution:
# template by 3xC and starkizard.
# contributors: 

#####################################################################################
from __future__ import division, print_function 

import sys
import os
from collections import Counter, deque, defaultdict
import itertools
import math
import io


"""Uncomment modules according to your need"""


# from bisect import bisect_left, bisect_right, insort
# from heapq import heappop, heapify, heappush
# from random import randint as rn
# from Queue import Queue as Q
# from copy import deepcopy
# from decimal import *
# import re
# import operator

#####################################################################################
# this enables you to write python3 code with PyPy2 (Python 2)
if sys.version_info[0] < 3:
    input = raw_input
    range = xrange

    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip

#####################################################################################

"""value of mod"""
MOD = 998244353
mod = 10**9 + 7

"""Uncomment next 4 lines if doing huge recursion"""
# import threading
# threading.stack_size(1<<27)
# sys.setrecursionlimit(10000

def prepare_factorial(mod=mod):
    """ returns two lists, factorial and inverse factorial modulo argument by default 10**9 +7 """
    # Comment code out when you don't need inverse factorial or vice versa
    fact = [1]
    for i in range(1, 200005):
        fact.append((fact[-1] * i) % mod)

    ifact = [0] * 200005
    ifact[200004] = pow(fact[200004], mod - 2, mod)
    for i in range(200004, 0, -1):
        ifact[i - 1] = (i * ifact[i]) % mod

    return fact, ifact

def modinv(n, p):
    """ returns N inverse modulo p """
    return pow(n, p - 2, p)

def ncr(n, r,  fact, ifact):
    """ takes 4 arguments: n , r and factorial and inverse factorial lists"""
    t = (fact[n] * (ifact[r]*ifact[n-r]) % MOD)% MOD
    return t


def get_n(Sum):    
    """this function returns the maximum n for which Summation(n) <= Sum"""
    ans = (-1 + sqrt(1 + 8*Sum))//2
    return ans

def sieve(n):
    """ returns a list of prime numbers till n """
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r



def divs(n, start=1):
    """ returns a list of all divisors till n """
    divisors = []
    for i in range(start, int(math.sqrt(n) + 1)):
        if n % i == 0:
            if n / i == i:
                divisors.append(i)
            else:
                divisors.extend([i, n // i])
    return divisors



def divn(n, primes):
    """ returns the number of divisors, two arguments n and the sieve till n """
    divs_number = 1
    for i in primes:
        if n == 1:
            return divs_number
        t = 1
        while n % i == 0:
            t += 1
            n //= i
        divs_number *= t



def lrfind(d, x, default=-1):
    """ Takes 2 arguments an iterable and an element. returns a tuple (firstoccurence,lastoccurence) -1 if not found """
    left = right = -1
    for i in range(len(d)):
        if d[i] == x:
            if left == -1: left = i
            right = i
    if left == -1:
        return default, default
    else:
        return left, right

def gcd(x, y): # math.gcd is slower
    """ returns greatest common divisor of x and y """
    while y:
        x, y = y, x % y
    return x

def check_sorted(a):
    ''' returns True/False '''
    for i in range(len(a)-1):
        if a[i]>a[i+1]:
            return False
    return True
def ceil(n, k=1): return n // k + (n % k != 0) #returns math.ceil but protecting against floating inconsistencies
def input(): return sys.stdin.readline().strip()
def ii(): return int(input()) #inputs integer
def mi(): return map(int, input().split()) # inputting space seperated variables for example x,y,z
def li(): return list(map(int, input().split())) #inputting a space seperated list of integers
def lw(): return input().split() #inputting a space seperated list of strings
def lcm(a, b): return abs(a * b) // gcd(a, b) #returns LCM of two arguments
def prr(a, sep=' ', end='\n'): print(sep.join(map(str, a)), end=end) #For printing an iterable with seperator sep as optional second argument (default : " "), ending character (default: "\n") as optional third
def dd(): return defaultdict(int) #returns a dictionary with values defaulted to 0
def ddl(): return defaultdict(list) #returns a dictionary with values defaulted to []
def write(s): return sys.stdout.write(s)












###################################################################
def main():
    #CODE GOES HERE:
    n=ii()
    v=li()
    v.sort()
    answer=v[0]
    for i in range(1,n):
        answer+=v[i]
        answer/=2
    print(answer)





















    
""" -------- Python 2 and 3 footer by Pajenegod and c1729 ---------"""

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192

class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)


class IOWrapper(IOBase):

    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s: self.buffer.write(s.encode('ascii'))
            self.read = lambda: self.buffer.read().decode('ascii')
            self.readline = lambda: self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

""" main function"""

if __name__ == '__main__':
    main()
    # threading.Thread(target=main).start()