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 <stdio.h>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <iostream>
#include<math.h>
#include<bitset>
#include<iomanip>
using namespace std;

int main() {
	int N; cin >> N;
	vector<long long>a(N),b(N);
	for (int i = 0; i < N; i++) { cin >> a[i]; }
	if (N % 2 == 0) {
		for (int i = 0; i < N / 2; i++) {
			b[i] = a[(N - 1) - 2 * i];
		}
		for (int i = N / 2; i < N; i++) {
			b[i] = a[2 * i-N];
		}
	}
	else {
		for(int i=0;i<=N/2;i++)b[i] = a[(N - 1) - 2 * i];
		for (int i = N / 2+1; i < N; i++) {
			b[i] = a[2 * i - N];
		}
	}
	for (int i = 0; i < N; i++)cout << b[i] << " ";
	
}