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)

#![allow(unused_imports)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;
use std::ops::Bound::*;

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

fn make_pairs(a: &Vec<usize>, n: usize) -> Vec<(usize, usize)> {
    let mut a_count = vec![0; n];
    for i in 0..n {
        a_count[a[i]] += 1;
    }
    let mut a_pairs = vec![];
    for i in 0..n {
        if a_count[i] > 0 {
            a_pairs.push((a_count[i], i));
        }
    }
    a_pairs.sort();
    a_pairs.reverse();
    a_pairs
}

fn make_count(a: &Vec<usize>, n: usize) -> Vec<usize> {
    let mut a_count = vec![0; n];
    for i in 0..n {
        a_count[a[i]] += 1;
    }
    a_count
}

fn main() {
    let n = read::<usize>();
    let a = read_vec::<usize>()
        .iter()
        .map(|&x| x - 1)
        .collect::<Vec<_>>();
    let b = read_vec::<usize>()
        .iter()
        .map(|&x| x - 1)
        .collect::<Vec<_>>();

    let a_count = make_count(&a, n);
    let b_count = make_count(&b, n);

    for i in 0..n {
        if a_count[i] + b_count[i] > n {
            println!("No");
            return;
        }
    }

    let a_pairs = make_pairs(&a, n);
    // debug!(a_pairs);
    let b_pairs = make_pairs(&b, n);
    let mut b_pairs = b_pairs.into_iter().collect::<BinaryHeap<_>>();

    let mut answers = HashMap::new();

    for (a_count, a_number) in a_pairs {
        let mut popped = vec![];
        let mut a_count_temp = a_count;
        while let Some((mut b_count, b_number)) = b_pairs.pop() {
            if a_number == b_number {
                popped.push((b_count, b_number));
                continue;
            }
            let reduced = min(a_count_temp, b_count);
            a_count_temp -= reduced;
            b_count -= reduced;
            if b_count > 0 {
                popped.push((b_count, b_number));
            }
            for _ in 0..reduced {
                answers.entry(a_number).or_insert(vec![]).push(b_number);
            }

            if a_count_temp == 0 {
                break;
            }
        }

        for elem in popped {
            b_pairs.push(elem);
        }

        if a_count_temp != 0 {
            println!("No");
            return;
        }
    }

    println!("Yes");
    let mut idx = vec![0; n];
    for anum in a {
        print!("{} ", answers[&anum][idx[anum]] + 1);
        idx[anum] += 1;
    }
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}