output_description: Print n integers in a line with spaces in between. The i-th integer should be
  b_i.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      n
      a_1 a_2 ... a_n
attempt: import sys
  
  sys.setrecursionlimit(10**6)
  from math import floor, ceil, sqrt, factorial, log
  from heapq import heappop, heappush, heappushpop
  from collections import Counter, defaultdict, deque
  from itertools import (
      accumulate,
      permutations,
      combinations,
      product,
      combinations_with_replacement,
  )
  from bisect import bisect_left, bisect_right
  from copy import deepcopy
  from operator import itemgetter
  from fractions import gcd
  
  mod = 10**9 + 7
  inf = float("inf")
  ninf = -float("inf")
  
  
  # 整数input
  def ii():
      return int(sys.stdin.readline().rstrip())  # int(input())
  
  
  def mii():
      return map(int, sys.stdin.readline().rstrip().split())
  
  
  def limii():
      return list(mii())  # list(map(int,input().split()))
  
  
  def lin(n: int):
      return [ii() for _ in range(n)]
  
  
  def llint(n: int):
      return [limii() for _ in range(n)]
  
  
  # 文字列input
  def ss():
      return sys.stdin.readline().rstrip()  # input()
  
  
  def mss():
      return sys.stdin.readline().rstrip().split()
  
  
  def limss():
      return list(mss())  # list(input().split())
  
  
  def lst(n: int):
      return [ss() for _ in range(n)]
  
  
  def llstr(n: int):
      return [limss() for _ in range(n)]
  
  
  # 本当に貪欲法か？ DP法では？？
  # 本当に貪欲法か？ DP法では？？
  # 本当に貪欲法か？ DP法では？？
  
  n = ii()
  arr = deque(limii())
  b = deque([])
  for i in range(n):
      if i % 2 == 1:
          x = arr.popleft()
          b.appendleft(str(x))
  
      else:
          x = arr.popleft()
          b.append(str(x))
  print(" ".join(list(b)))
  
problem_description: 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:
  
    1. Append a_i to the end of b.
    2. Reverse the order of the elements in b.
  
  Find the sequence b obtained after these n operations.
samples: [{"input": "4\n    1 2 3 4", "output": "4 2 1 3\n    \n\n  * After step 1 of the first operation, b becomes: 1.\n  * After step 2 of the first operation, b becomes: 1.\n  * After step 1 of the second operation, b becomes: 1, 2.\n  * After step 2 of the second operation, b becomes: 2, 1.\n  * After step 1 of the third operation, b becomes: 2, 1, 3.\n  * After step 2 of the third operation, b becomes: 3, 1, 2.\n  * After step 1 of the fourth operation, b becomes: 3, 1, 2, 4.\n  * After step 2 of the fourth operation, b becomes: 4, 2, 1, 3.\n\nThus, the answer is `4 2 1 3`.\n\n* * *"}, {"input": "3\n    1 2 3", "output": "3 1 2\n    \n\nAs shown above in Sample Output 1, b becomes 3, 1, 2 after step 2 of the third\noperation. Thus, the answer is `3 1 2`.\n\n* * *"}, {"input": "1\n    1000000000", "output": "1000000000\n    \n\n* * *"}, {"input": "6\n    0 6 7 6 7 0", "output": "0 6 6 0 7 7"}]