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 Ruby (2.3.3) (judge verdict: Wrong Answer)

a = gets.strip.to_i
arr = gets.strip.split(' ').map(&:to_i)

map = {}
appeared = []
f = []
arr.each_with_index do |a, index|
  f.push([])
  if map.has_key?(a)
    appeared.push(map[a])
  else
    appeared.push(false)
    map[a] = index
  end
end

def mod(k)
  k % 1000000007
end
#f[0].push(1)
(0...arr.size()).each do |i|
  (0...arr.size()).each do |j|
    if i == j
      f[i].push(1)
    elsif j < i
      f[i].push(0)
    else
      #print("#{i} #{j}\n")
      if i == 0
        if appeared[j]
          f[i].push(f[i][j-1])
        else
          f[i].push(mod(f[i][j-1] + 1))
        end
      else
        if appeared[j]
          f[i].push(mod(f[i-1][j-1]+ f[i][j-1] - f[i][appeared[j]]))
        else
          f[i].push(mod(f[i-1][j-1]+ f[i][j-1]))
        end
      end
    end
  end
end

(0...arr.size()).each do |i|
  puts f[i][-1]
end