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>
using namespace std;
int a[200000], b[200000];
int main()
{
	int n;
	cin >> n;
	if (n < 20000)
	{
		for (int i = 1; i <=n; i++)
			cin >> a[i];
		for (int j = 1; j <=n; j++)
		{
			b[j] = a[j];
			if (j > 1)
			{
				int temp;
				int m = (j + 1) / 2;
				if (j % 2)
				{
					for (int i = 0;; i++)
					{
						temp = b[m + i + 1];
						b[m + i + 1] = b[m - i -1];
						b[m - i -1] = temp;
						if (m - i -1 == 1)
							break;
					}
				}
			    else
				{
					for (int i = 1;; i++)
					{
						temp = b[m + i ];
						b[m + i] = b[m - i + 1];
						b[m - i + 1] = temp;
						if (m - i + 1 == 1)
							break;
					}
				}
			}
		}
	}
	for (int i = 1; i <= n; i++)
		cout << b[i] << " ";
	return 0;
}