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

int main(void){
    int n,i;
    cin >> n;

    vector<int>  a(n+1);
    for(i=1; i<=n; i++){
        cin >> a.at(i);
    }


    if(n % 2 == 0  && n > 2){
        for(i=n; i>=2; i-=2){
            cout << a.at(i) << " ";
        }
        for(i=1; i<n-1; i+=2){
            cout << a.at(i) << " ";
        }
        cout << a.at(n-1) << endl;
    }else if(n % 2 == 1 && n > 1){
        for(i=n; i>=1; i-=2){
            cout << a.at(i) << " ";
        }
        for(i=2; i<n-1; i+=2){
            cout << a.at(i) << " ";
        }
        cout << a.at(n-1) << endl;
    }



    if(n == 1){
        cout << a.at(1) << endl;
    }

    if(n == 2){
        cout << a.at(2) << " " <<a.at(1) << endl;
    }


    return 0;
}