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: Wrong Answer)

#include<iostream>
#include<string>
#include<map>
#include<vector>
using namespace std;
int main()
{
	int g;
	int n;
	cin >> n;
	vector<int>a;
	for (int i = 0; i < n; i++) {
		cin >> g;
		a.push_back(g);
	}
	int k;
	k = n;
	if (n % 2 == 0) {
	
		for (int i = 0; i < n/2; i++) {
			cout << a[k-1]<<" ";
			k -=2;

		}
		for (int i = 0; i < n / 2; i++) {
			cout << a[i*2]<<" ";

		}
	}
	else if(n%4==1){
	
		for (int i = 0; i < n / 2; i++) {
			cout << a[k - 2] << " ";
			k -= 2;

		}
		for (int i = 0; i < n / 2+1; i++) {
			cout << a[i * 2] << " ";

		}
	}
	else {


		for (int i = 0; i < n / 2+1; i++) {
			cout << a[k - 1] << " ";
			k -= 2;

		}
		for (int i = 0; i < n / 2 ; i++) {
			cout << a[i * 2+1] << " ";

		}
	}

	

		
}