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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <cstdlib>
#include <string>
#include <vector>
#include <algorithm>
#include <queue>
#define MAXN 200000;
#define Amari 1000000007;
#define p pair <int,int>
using namespace std;
int main(int argc,char const* argv[])
{
  int n;
  long long temp;
  cin >> n;
  vector<long long> ans(n,0);
  if(n % 2 ==0)
    {
      for(int i=0;i<n;i++)
	{
	  cin >> temp;
	  if(i==0)
	    {
	      ans[n/2]=temp;
	    }
	  else
	    {
	      if(i % 2 == 0) ans[n/2+i/2]=temp;
	      else if(i % 2 !=0) ans[n/2-((i+1)/2)]=temp;
	    }
	}
    }
  else
    {
       for(int i=0;i<n;i++)
	{
	  cin >> temp;
	  if(i==0)
	    {
	      ans[n/2]=temp;
	    }
	  else
	    {
	      if(i % 2 != 0) ans[n/2+((i+1)/2)]=temp;
	      else if(i % 2 ==0) ans[n/2-i/2]=temp;
	    }
	}
    }
  for(int i=0;i<n;i++)
    {
      cout << ans[i] << " ";
    }
  return 0;
}