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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;


int main()
{
    int n;
    cin>>n;
    vector<int> v;
    
    for(int i=0;i<n;i++)
    {
        
        int x;
        cin>>x;
        v.push_back(x);
    }
    vector<int> ans(n);

     if(n%2==0)
     {
          int mid=n/2;


          for(int i=0;i<n;i+=2)
          {
              int in=i/2;
              ans[mid+in]=v[i];
              ans[mid-in-1]=v[i+1];
          }   


          for(auto i:ans)
          cout<<i<<" ";
          cout<<endl;
     }
     else
     {

                ans[n/2]=v[0];
         for(int i=1;i<n;i+=2)
         {
            ans[n/2+(i+1)/2]=v[i];
            ans[n/2-(i+1)/2]=v[i+1];
            
         }
         for(auto i:ans)
         cout<<i<<" ";
         cout<<endl;

     }
}