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: Runtime Error)

import java.util.*;

public class Main{
    public static long mod = 1000000007;
    public static Scanner sc = new Scanner(System.in);
    public static void main(String args[]){

        int N = sc.nextInt();

        int[] t = new int[N*2];
        int l = 0;
        int r = 1;
        int k = 1;
        int ii = 1;
        for( int i = 0; i < N; i++, k++ ){
            if( (k & 1) == 0 ){
                t[N+l] = sc.nextInt();
                l--;
            }else{
                t[N+r] = sc.nextInt();
                r++;
            }
            // if( ii++ == 3) { k++; ii=0; }
        }

        int ind = 1;
        if( k%2 == 1 ){
        	for( int i = N+l+1; i < N+r; i++ ) {
        		System.out.printf( "%d%c", t[i], ind == N ? '\n' : ' ' );
        	}
        }else{
        	for( int i = N+r-1; i > N+l; i-- ) {
        		System.out.printf( "%d%c", t[i], ind == N ? '\n' : ' ' );
        	}
        }


    }
}