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 itertools::Itertools;
use proconio::input;

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

    if let Some(ans) = solve(n, &a, &b) {
        println!("Yes");
        println!("{}", ans.iter().map(|&v| v.to_string()).join(" "))
    } else {
        println!("No")
    }
}

fn solve(n: usize, a: &[usize], b: &[usize]) -> Option<Vec<usize>> {
    let mut d = 0;
    let b = [b; 2].concat();
    for i in (0..n).rev() {
        while a[i] == b[i + d] {
            d += 1;
            if d >= n {
                return None;
            }
        }
    }
    if (0..n).any(|i| a[i] == b[i + d]) {
        return None;
    }
    Some((&b[d..d + n]).into())
}