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.io.*;
import java.math.*;
import java.util.*;

public class Main {
    private static boolean debug = false;
    private static boolean elapsed = true;

    private static PrintWriter _out = new PrintWriter(System.out);
    private static PrintWriter _err = new PrintWriter(System.err);

    private void solve(Scanner sc) {
        int n = sc.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; ++i) {
            a[i] = sc.nextInt();
        }

        boolean flg = true;
        int[] b = new int[n];
        int left = n / 2;
        int right = left - 1;
        for (int i = 0; i < n; ++i) {
            if (flg) {
                b[++right] = a[i];
            } else {
                b[--left] = a[i];
            }
            flg = !flg;
        }
        if (flg) {
            for (int i = 0; i < n; ++i) {
                if (i > 0) {
                    _out.print(" ");
                }
                _out.print(b[i]);
            }
        } else {
            for (int i = n - 1; i >= 0; --i) {
                _out.print(b[i]);
                if (i > 0) {
                    _out.print(" ");
                }
            }
        }
        _out.println();
    }
    private static BigInteger C(long n, long r) {
        BigInteger res = BigInteger.ONE;
        for (long i = n; i > n - r; --i) {
            res = res.multiply(BigInteger.valueOf(i));
        }
        for (long i = r; i > 1; --i) {
            res = res.divide(BigInteger.valueOf(i));
        }
        return res;
    }
    private static BigInteger P(long n, long r) {
        BigInteger res = BigInteger.ONE;
        for (long i = n; i > n - r; --i) {
            res = res.multiply(BigInteger.valueOf(i));
        }
        return res;
    }
    /*
     * 10^10 > Integer.MAX_VALUE = 2147483647 > 10^9
     * 10^19 > Long.MAX_VALUE = 9223372036854775807L > 10^18
     */
    public static void main(String[] args) {
        long S = System.currentTimeMillis();

        Scanner sc = new Scanner(System.in);
        new Main().solve(sc);
        _out.flush();

        long G = System.currentTimeMillis();
        if (elapsed) {
            _err.println((G - S) + "ms");
        }
        _err.flush();
    }
}