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, unused_macros)]
use std::io::Write;
use std::collections::{VecDeque, HashMap, BinaryHeap, BTreeMap};

struct IO<'a> {
    sw: std::str::SplitWhitespace<'a>,
    out: Vec<u8>,
}

impl IO<'_> {
    pub fn read<T>(&mut self) -> T where
        T: std::str::FromStr,
        <T as std::str::FromStr>::Err: std::fmt::Debug {
        self.sw.next().unwrap().parse().unwrap()
    }
}

impl std::io::Write for IO<'_>  {
    fn write(&mut self, buf: &[u8]) -> std::io::Result<usize> {
        std::io::Write::write(&mut self.out, buf)
    }
    fn flush(&mut self) -> std::io::Result<()> {
        std::io::Write::flush(&mut self.out)
    }
}

macro_rules! write_vec {
    ($io:expr, $v:tt) => {
        for (i, _v) in $v.iter().enumerate() {
            write!($io, "{:?}{}", _v, if i == $v.len()-1 {"\n"} else {" "})?;
        }
    };
}

macro_rules! write_rep {
    ($io:expr, $v:tt) => (
        write!($io, "{:?}\n", $v)?;
    );
    ($io:expr, $v:tt, $($arg:tt),+) => (
        write!($io, "{:?} ", $v)?;
        write_rep!($io, $($arg),+);
    );
}

fn main() -> Result<(), std::io::Error> {
    let mut buf = String::new();
    std::io::Read::read_to_string(&mut std::io::stdin(), &mut buf)?;
    let sw = buf.split_whitespace();
    let mut io = IO { sw, out: Vec::new() };
    solve(&mut io)?;
    std::io::Write::write_all(&mut std::io::stdout(), &io.out)?;
    Ok(())
}

// AtCoder Beginner Contest 178 F - Contrast
fn solve(io: &mut IO) -> Result<(), std::io::Error> {
    let n: usize = io.read();
    let mut a_cnt: Vec<usize> = vec![0; n+1];
    for _ in 0..n {
        let a: usize = io.read();
        a_cnt[a] += 1;
    }
    let mut b_vec: Vec<usize> = vec![0; n];
    let mut b_cnt: Vec<usize> = vec![0; n+1];
    
    for i in 0..n {
        let b = io.read();
        b_vec[i] = b;
        b_cnt[b] += 1;
    }
    for i in 1..=n {
        if a_cnt[i]+b_cnt[i] > n {
            writeln!(io, "No")?;
            return Ok(())
        }
    }
    let mut cap: usize = 0;
    for i in 0..n {
        a_cnt[i+1] += a_cnt[i];
        b_cnt[i+1] += b_cnt[i];
        cap = cap.max(a_cnt[i+1].saturating_sub(b_cnt[i]))
    }
    writeln!(io, "Yes")?;
    for i in 0..n {
        write!(io, "{}", b_vec[(i+cap)%n])?;
        if i == n-1 {
            writeln!(io)?;
        } else {
            write!(io, " ")?;
        }
    }
    Ok(())
}