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 Rust (1.15.1) (judge verdict: Accepted)

#![allow(non_snake_case)]
#![allow(unused_variables)]

fn main() {
    let n: usize = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().parse().unwrap()
    };
    let a: Vec<i64> = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.split_whitespace()
            .map(|x| x.parse().unwrap())
            .collect()
    };

    let ans = if n % 2 == 0 {
        let L = (0..(n / 2)).map(|i| a[2 * (n / 2 - i - 1) + 1].to_string()).collect::<Vec<_>>().join(" ");
        let R = (0..(n / 2)).map(|i| a[2 * i].to_string()).collect::<Vec<_>>().join(" ");
        format!("{} {}", L, R)
    } else {
        let L = (0..(n / 2 + 1)).map(|i| a[2 * (n / 2 - i)].to_string()).collect::<Vec<_>>().join(" ");
        let R = (0..(n / 2)).map(|i| a[2 * i + 1].to_string()).collect::<Vec<_>>().join(" ");
        format!("{} {}", L, R)
    };

    println!("{}", ans);
}