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: Wrong Answer)

use std::collections::BTreeMap;

use proconio::input;

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

    let mut acount = BTreeMap::<usize, usize>::new();
    for &ai in &a {
        *acount.entry(ai).or_insert(0) += 1;
    }
    let acount = acount;

    let mut bcount = BTreeMap::<usize, usize>::new();
    for &bi in &b {
        *bcount.entry(bi).or_insert(0) += 1;
    }
    let bcount = bcount;

    for ai in &a {
        if acount.get(ai).unwrap_or(&0) + bcount.get(ai).unwrap_or(&0) > n {
            println!("No");
            return;
        }
    }

    let mut index: Vec<Option<usize>> = vec![None; n + 1];
    for i in 0..n {
        index[a[i]] = Some((i + 1) % n);
    }

    let mut rem = Vec::<usize>::with_capacity(n);
    let mut res = vec![None; n];
    for &bi in &b {
        let s = index[bi];
        if s.is_none() {
            for _ in 0..bcount[&bi] {
                rem.push(bi);
            }
            continue;
        }
        let s = s.unwrap();
        let mut r = bcount[&bi];
        for c in 0..bcount[&bi] {
            if res[(s + c) % n].is_some() {
                break;
            }
            r -= 1;
            res[(s + c) % n] = Some(bi);
        }
        for _ in 0..r {
            rem.push(bi);
        }
    }

    for i in 0..n {
        if res[i].is_none() {
            res[i] = Some(rem.pop().unwrap());
        }
    }

    println!("Yes");
    for i in 0..n {
        print!("{}{}", res[i].unwrap(), if i + 1 < n { ' ' } else { '\n' });
    }
}