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)

fn main() {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    let n: usize = s.trim().parse().unwrap();
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    let a: Vec<usize> = s.trim().split_whitespace().map(|x| x.parse().unwrap()).collect();
    let mut b: Vec<usize> = vec![0; n];
    let z: usize = n/2 - (if n%2 == 0 {1} else {0});
    let mut i = 0;
    let mut j = 0;
    loop {
        b[z-j] = a[i];
        i += 1;
        if i >= n {break;}
        b[z+1+j] = a[i];
        i += 1;
        j += 1;
        if i >= n {break;}
    }
    if n%2 == 0 {b.reverse()}
    for i in 0..n {
        print!("{}{}", b[i], if i == n - 1 {"\n"} else {" "});
    }
}