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)

fn main() {
    let n: usize = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        buf.trim_end().parse().unwrap()
    };
    let a: Vec<usize> = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        let iter = buf.split_whitespace();
        iter.map(|x| x.parse().unwrap()).collect()
    };
    let b: Vec<usize> = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        let iter = buf.split_whitespace();
        iter.map(|x| x.parse().unwrap()).collect()
    };

    let mut c = vec![0; n + 1];
    let mut d = vec![0; n + 1];
    for i in 0..n {
        c[a[i]] += 1;
        d[b[i]] += 1;
    }

    for i in 1..=n {
        if c[i] + d[i] > n {
            return println!("No");
        }
        c[i] += c[i - 1];
        d[i] += d[i - 1];
    }
    println!("Yes");
    let x = (1..=n).map(|i| c[i] - d[i - 1]).max().unwrap();
    println!(
        "{}",
        (0..n).map(|i| b[(i + n - x) % n].to_string())
            .collect::<Vec<_>>()
            .join(" ")
    );
}