Problem:
Score : 600 points

Problem Statement

Given are two sequences A and B, both of length N.
A and B are each sorted in the ascending order.
Check if it is possible to reorder the terms of B so that for each i (1 \leq i \leq N) A_i \neq B_i holds, and if it is possible, output any of the reorderings that achieve it.

Constraints

1\leq N \leq 2 \times 10^5

1\leq A_i,B_i \leq N

A and B are each sorted in the ascending order.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \cdots A_N
B_1 B_2 \cdots B_N

Output

If there exist no reorderings that satisfy the condition, print No.

If there exists a reordering that satisfies the condition, print Yes on the first line.
After that, print a reordering of B on the second line, separating terms with a whitespace.

If there are multiple reorderings that satisfy the condition, you can print any of them.

Sample Input 1

6
1 1 1 2 2 3
1 1 1 2 2 3

Sample Output 1

Yes
2 2 3 1 1 1

Sample Input 2

3
1 1 2
1 1 3

Sample Output 2

No

Sample Input 3

4
1 1 2 3
1 2 3 3

Sample Output 3

Yes
3 3 1 2

Solution in Rust (1.42.0) (judge verdict: Time Limit Exceeded)

#[allow(unused_imports)]
use {
    itertools::Itertools,
    proconio::{fastout, input, marker::*},
    std::cmp::*,
    std::collections::*,
    std::io::Write,
    std::ops::*,
};

macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

#[fastout]
fn main() {
    input! {
        n: usize,
        a: [usize; n],
        b: [usize; n],
    }

    let mut map = BTreeMap::new();
    for &a in a.iter() {
        *map.entry(a).or_insert(0) += 1;
    }
    for &a in b.iter() {
        *map.entry(a).or_insert(0) += 1;
    }
    for &k in map.values() {
        if k > n {
            println!("No");
            return;
        }
    }

    let mut q = std::collections::VecDeque::<_>::new();
    for i in 0..n {
        q.push_back((i, a[i]));
    }
    let mut ans = vec![0; n];
    for &b in b.iter() {
        loop {
            if let Some((i, a)) = q.pop_front() {
                if a != b {
                    ans[i] = b;
                    break;
                }
                q.push_back((i, a));
            }
        }
    }
    println!("Yes");
    println!("{}", ans.iter().join(" "));
}