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: Accepted)

#![allow(clippy::needless_range_loop)]
#![allow(unused_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

use itertools::Itertools;
use proconio::input;
use proconio::marker::*;

fn main() {
    input! {
        n: usize,
        a: [Usize1; n],
        b: [Usize1; n],
    }

    let mut cnt_a = vec![0; n];
    let mut cnt_b = vec![0; n];

    for &x in &a {
        cnt_a[x] += 1;
    }

    for &x in &b {
        cnt_b[x] += 1;
    }

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

    let mut a_r = vec![None; n];
    let mut b_l = vec![None; n];

    for (index, &x) in a.iter().enumerate() {
        a_r[x] = Some(index);
    }

    for (index, &x) in b.iter().enumerate() {
        if b_l[x].is_none() {
            b_l[x] = Some(index);
        }
    }

    let rot = a_r
        .iter()
        .zip(b_l.iter())
        .map(|(&a, &b)| {
            if let Some(r) = a {
                if let Some(l) = b {
                    if r >= l {
                        return r - l + 1;
                    }
                }
            }
            0
        })
        .max()
        .unwrap();

    
    println!("Yes");

    for i in 0..n {
        let d = if i == n - 1 {
            "\n"
        } else {
            " "
        };
        print!("{}{}", b[(i + n - rot) % n] + 1, d);
    }
}