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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.Scanner;

public class Main {
	static ArrayList<Long> b = new ArrayList<>();
	
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		long[] a = new long[n];
		for (int i = 0; i < n; i++) {
			a[i] = scan.nextInt();
		}
		if (n % 2 == 0) {
			epush(a, n);			
		} else {
			opush(a, n);
		}
		for (int i = 0; i < n; i++) {
			System.out.print(b.get(i));
			System.out.print(" ");
		}
		System.out.println("");
	}
	public static void epush(long a[], int n) {
		for (int i = n-1; i >= 0; i--) {
			if (i % 2 == 1) {
				b.add(a[i]);
			}
		}
		for (int i = 0; i < n; i++) {
			if (i % 2 == 0) {
				b.add(a[i]);
			}
		}
		return;
	}
	public static void opush(long a[], int n) {
		for (int i = n-1; i >= 0; i--) {
			if (i % 2 == 0) {
				b.add(a[i]);
			}
		}
		for (int i = 0; i < n; i++) {
			if (i % 2 == 1) {
				b.add(a[i]);
			}
		}
		return;
	}
}