Problem:
Score : 300 points

Problem Statement

You are given an integer sequence of length n, a_1, ..., a_n.
Let us consider performing the following n operations on an empty sequence b.

The i-th operation is as follows:

Append a_i to the end of b.

Reverse the order of the elements in b.

Find the sequence b obtained after these n operations.

Constraints

1 \leq n \leq 2\times 10^5

0 \leq a_i \leq 10^9

n and a_i are integers.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_n

Output

Print n integers in a line with spaces in between.
The i-th integer should be b_i.

Sample Input 1

4
1 2 3 4

Sample Output 1

4 2 1 3

After step 1 of the first operation, b becomes: 1.

After step 2 of the first operation, b becomes: 1.

After step 1 of the second operation, b becomes: 1, 2.

After step 2 of the second operation, b becomes: 2, 1.

After step 1 of the third operation, b becomes: 2, 1, 3.

After step 2 of the third operation, b becomes: 3, 1, 2.

After step 1 of the fourth operation, b becomes: 3, 1, 2, 4.

After step 2 of the fourth operation, b becomes: 4, 2, 1, 3.

Thus, the answer is 4 2 1 3.

Sample Input 2

3
1 2 3

Sample Output 2

3 1 2

As shown above in Sample Output 1, b becomes 3, 1, 2 after step 2 of the third operation. Thus, the answer is 3 1 2.

Sample Input 3

1
1000000000

Sample Output 3

1000000000

Sample Input 4

6
0 6 7 6 7 0

Sample Output 4

0 6 6 0 7 7

Solution in Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

import Control.Applicative
import Control.Monad
import Data.List
import Data.Char
strToInt s = (read :: String -> Int) s
strToInteger s = (read :: String -> Integer) s
strToDouble s = (read :: String -> Double) s

main = do
    -- 整数の入力
    n <- readLn
    -- スペース区切り整数の入力
    l <- map strToInt . words <$> getLine
    -- 出力
    forM (intersperse ' ' . concat . map show $ func $ solve n l []) putChar

func :: [a] -> [a]
func l = if odd $ length l then reverse l else id l

solve :: Int -> [Int] -> [Int] -> [Int]
solve n (a:as) b
    | null as = a:b
    | even n = solve (pred n) as (b ++ [a])
    | otherwise = solve (pred n) as (a:b)