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 n;
int main() {
    cin >> n;
    vector<long long> a(n, 0);
    for(int i = 0; i < n; ++i) cin >> a.at(i);

    list<long long> b;
    int init = n-1;

    if(init % 2 != 0) {
        for(int i = 0; i < n; ++i) {
            if(i % 2 == 0) {
                b.push_back(a.at(i));
            }else {
                b.push_front(a.at(i));
            }
        }
    }else {
        b.push_back(a.at(0));
        for(int i = 1; i < n; ++i) {
            if(i % 2 != 0) {
                b.push_back(a.at(i));
            }else {
                b.push_front(a.at(i));
            }
        }
    }

    for(int i = 0; i < n; ++i) {
        cout << b.front() << " ";
        b.pop_front();
    }
}