Problem:
Score : 300 points

Problem Statement

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}.

Sample Input 1

2
3 4

Sample Output 1

3.5

If you start with two ingredients, the only choice is to put both of them in the pot. The value of the ingredient resulting from the ingredients of values 3 and 4 is (3 + 4) / 2 = 3.5.

Printing 3.50001, 3.49999, and so on will also be accepted.

Sample Input 2

3
500 300 200

Sample Output 2

375

You start with three ingredients this time, and you can choose what to use in the first composition. There are three possible choices:

Use the ingredients of values 500 and 300 to produce an ingredient of value (500 + 300) / 2 = 400. The next composition will use this ingredient and the ingredient of value 200, resulting in an ingredient of value (400 + 200) / 2 = 300.

Use the ingredients of values 500 and 200 to produce an ingredient of value (500 + 200) / 2 = 350. The next composition will use this ingredient and the ingredient of value 300, resulting in an ingredient of value (350 + 300) / 2 = 325.

Use the ingredients of values 300 and 200 to produce an ingredient of value (300 + 200) / 2 = 250. The next composition will use this ingredient and the ingredient of value 500, resulting in an ingredient of value (250 + 500) / 2 = 375.

Thus, the maximum possible value of the last ingredient remaining is 375.

Printing 375.0 and so on will also be accepted.

Sample Input 3

5
138 138 138 138 138

Sample Output 3

138

Solution in Haskell (GHC 7.10.3) (judge verdict: Accepted)

--{-# LANGUAGE BangPatterns #-}
--{-# LANGUAGE ScopedTypeVariables #-}
--{-# LANGUAGE FlexibleContexts #-}
--{-# LANGUAGE ViewPatterns #-}
--{-# LANGUAGE OverloadedStrings #-}

import qualified Data.Vector as V
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import qualified Data.ByteString.Char8 as BC
import qualified Data.Map as MP
import qualified Data.Set as ST
import qualified Data.Sequence as SQ
import Data.Array
import Data.Array.Unboxed
import Data.Array.ST
import Data.Array.IO
import Data.Maybe
import Data.Ord
import Data.Ratio
import Data.Char
import Data.Bits
import Data.List
import Data.Tuple
import Data.Functor
import Data.IORef
import Data.STRef
import Control.Applicative
import Control.Monad
import Control.Monad.ST
import Debug.Trace
-- import qualified Data.Vector.Algorithms.Merge as VAM
-- import qualified Data.Heap as HP
-- import Numeric.Extra
-- import Data.Tuple.Extra
-- import Data.List.Extra

--------------------------------------------------------------------------
solve [x] = x
solve (x:y:xs) = solve (((x+y)/2):xs)

main = do
    n <- int
    xs <- sort <$> sLineToDoubleL
    print $ solve xs
--------------------------------------------------------------------------

-- Input Util
int :: IO Int
int = readLn 

double :: IO Double
double = readLn 

str :: IO String
str = getLine

strBS :: IO BC.ByteString
strBS = BC.getLine

bsToInt :: BC.ByteString -> Int
bsToInt = fst . fromJust . BC.readInt

bsToDouble :: BC.ByteString -> Double
bsToDouble = read . BC.unpack

parseInt :: BC.ByteString -> Maybe (Int, BC.ByteString)
parseInt = BC.readInt . BC.dropWhile isSpace

sLineToIntL :: IO [Int]
sLineToIntL = strBS >>= return . map bsToInt . BC.words

sLineToDoubleL :: IO [Double]
sLineToDoubleL = strBS >>= return . map bsToDouble . BC.words

sLineToIntV :: Int -> IO (VU.Vector Int)
sLineToIntV n = strBS >>= return . VU.unfoldrN n parseInt

sLineToDoubleV :: IO (VU.Vector Double)
sLineToDoubleV = strBS >>= return . VU.fromList . map bsToDouble . BC.words

mLinesToIntL :: Int -> IO [Int]
mLinesToIntL n = replicateM n (strBS >>= return . bsToInt)

mLinesToDoubleL :: Int -> IO [Double]
mLinesToDoubleL n = replicateM n (strBS >>= return . bsToDouble)

mLinesToIntV :: Int -> IO (VU.Vector Int)
mLinesToIntV n = VU.replicateM n (strBS >>= return . bsToInt)

mLinesToDoubleV :: Int -> IO (VU.Vector Double)
mLinesToDoubleV n = VU.replicateM n (strBS >>= return . bsToDouble)

mLinesToTupleL :: Int -> IO [(Int, Int)]
mLinesToTupleL n = replicateM n (strBS >>= return . (\[a,b] -> (a,b)) . map bsToInt . BC.words)

mLinesToTripleL :: Int -> IO [(Int, Int, Int)]
mLinesToTripleL n = replicateM n (strBS >>= return . (\[a,b,c] -> (a,b,c)) . map bsToInt . BC.words)

mLinesToTupleV :: Int -> IO (VU.Vector (Int, Int))
mLinesToTupleV n = VU.replicateM n $ do
    bs <- strBS
    let
        Just (a, bs') = parseInt bs
        Just (b, _) = parseInt bs'
    return (a,b)
    
mLinesToTripleV :: Int -> IO (VU.Vector (Int, Int, Int))
mLinesToTripleV n = VU.replicateM n $ do
    bs <- strBS
    let
        Just (a, bs') = parseInt bs
        Just (b, bs'') = parseInt bs'
        Just (c, _) = parseInt bs''
    return (a,b,c)

-- output
yesnoL :: Bool -> IO ()
yesnoL cond
    | cond = putStrLn "Yes"
    | otherwise = putStrLn "No"

yesnoU :: Bool -> IO ()
yesnoU cond
    | cond = putStrLn "YES"
    | otherwise = putStrLn "NO"

-- others
strToChr :: String -> Char
strToChr [x] = x

chrToStr :: Char -> String
chrToStr x = [x]

m1 :: Int -> Int
m1 x = subtract 1 x

toDouble :: Int -> Double
toDouble x = fromIntegral x

toInt :: Double -> Int
toInt x = floor x