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: Runtime Error)

fn f(from: i32, to: i32, m: i32) -> i32 {
    if from <= to {
        to - from
    } else {
        to + m - from
    }
}
fn compute(a: &Vec<i32>, base: i32, m: i32) -> i64 {
    let mut ans: i64 = 0;
    for i in 0..(a.len()-1) {
        let (x, y) = (a[i], a[i+1]);
        let (a1, a2) = (f(x, y, m), 1+f(base, y, m));
        ans += min(a1, a2) as i64;
    }
    ans
}
fn main() {
    let (_, m): (i32, i32) = readln();
    let a_: Vec<i32> = readln();
    let a: Vec<i32> = a_.iter().map(|x| x-1).collect();
    let mut d = 1;
    while d * 2 <= m {
         d *= 2;
    }
    let mut cur = (12 * a[0] + 3) % m;
    let mut answer: i64 = i64::max_value();
    loop {
        let (x, y, z) = (
            compute(&a, (cur+m-1)%m, m),
            compute(&a, cur, m),
            compute(&a, (cur+1)%m, m)
        );
        answer = min(min(answer, x), min(y, z));
        if x >= y && y <= z {
            break
        } else if y - x > y - z {
            // go to left
            cur = (cur + m - d) % m;
        } else {
            // go to right
            cur = (cur + m + d) % m;
        }
        if d != 1 {
            d /= 2;
        }
    }
    println!("{}", answer);
}

// --- template ---
#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::HashSet;

pub trait FromLn {
    fn fromln(s: &str) -> Self;
}
pub fn readln<T: FromLn>() -> T {
    let mut buf = String::new();
    let _ = ::std::io::stdin().read_line(&mut buf).unwrap();
    T::fromln(buf.trim())
}
pub fn readlns<T: FromLn>(n: usize) -> Vec<T> {
    let mut vs = vec![];
    for _ in 0..n { vs.push(readln()); }
    vs
}
macro_rules! fromln_primitives {
    ($($t:ty),*) => { $(
        impl FromLn for $t {
            fn fromln(s: &str) -> $t {
                s.parse().unwrap()
            }
        }
    )* }
}
fromln_primitives!(String, bool, f32, f64, isize, i8, i16, i32, i64, usize, u8, u16, u32, u64);
impl<T> FromLn for Vec<T> where T: FromLn {
    fn fromln(s: &str) -> Vec<T> {
        s.split_whitespace().map(T::fromln).collect()
    }
}
impl FromLn for Vec<char> {
    fn fromln(s: &str) -> Vec<char> {
        s.chars().collect()
    }
}
macro_rules! fromln_tuple {
    ($($t:ident),*) => {
        impl<$($t),*> FromLn for ($($t),*) where $($t: FromLn),* {
            fn fromln(s: &str) -> ($($t),*) {
                let mut it = s.split_whitespace();
                let t = ($($t::fromln(it.next().unwrap())),*);
                assert_eq!(it.next(), None);
                t
            }
        }
    }
}
fromln_tuple!(A, B);
fromln_tuple!(A, B, C);
fromln_tuple!(A, B, C, D);
fromln_tuple!(A, B, C, D, E);
fromln_tuple!(A, B, C, D, E, F);