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)

-- Question C

import Control.Applicative
import Control.Monad

main = do
  n <- readLn ::IO Int
  zs <- (map read . words) <$> getLine 
  putStr $ unwords ( map show ( try' n zs) )
  
try' n zs = if n `mod` 2 == 0
                then (reverse (try1 1 n [] zs)) ++ try2 2 n [] zs
                else (reverse (try2 2 n [] zs)) ++ try1 1 n [] zs				

try1 :: Int -> Int -> [Int] -> [Int] -> [Int]
try1 i k ys xs  
  | i > k = ys 
  | otherwise = try1 (i+2) k ( ys ++ [xs !! (i-1)] ) xs
  
try2 :: Int -> Int -> [Int] -> [Int] -> [Int]
try2 i k ys xs  
  | i > k = ys 
  | otherwise = try2 (i+2) k ( ys ++ [xs !! (i-1)] ) xs