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: Time Limit Exceeded)

import java.util.*;
import java.lang.*;
public class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);

		int N = sc.nextInt();
		long[] ans = new long[N+1];
		int b = N/2+1;
		int c = N/2+2;
		int d = N/2;
		if( N%2!=0 ){
			for( int i=1; i<=N; i++ ){
				if( i%2!=0 ){
					ans[b] = Long.parseLong(sc.next());
					b--;
				}else{
					ans[c] = Long.parseLong(sc.next());
					c++;
				}
			}
		}
		if( N%2==0 ){
			for( int i=1; i<=N; i++ ){          
				if( i%2!=0 ){
					ans[b] = Long.parseLong(sc.next());
					b++;
				}else{
					ans[d] = Long.parseLong(sc.next());
					d--;
				}	
            }              
		}
		String ss = "A";
		for( int i=1; i<=N; i++ ){
			ss = ss + " " + String.valueOf(ans[i]);
		}
		System.out.println(ss.substring(2));
	}
}