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

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let sw = buf.split_whitespace();
    let mut io = IO { sw, out: Vec::new() };
    solve(&mut io);
    std::io::stdout().write_all(&io.out).unwrap();
}

// AtCoder Beginner Contest 178 F - Contrast
fn solve(io: &mut IO) {
    let n: usize = io.read();
    let mut am: BTreeMap<usize, usize> = BTreeMap::new();
    for _ in 0..n {
        *am.entry(io.read()).or_default() += 1;
    }
    let mut bm: BTreeMap<usize, usize> = BTreeMap::new();
    for _ in 0..n {
        *bm.entry(io.read()).or_default() += 1;
    }
    let mut bh: BinaryHeap<(usize, usize)> = BinaryHeap::from(bm.iter().map(
        |(&k, &v)|{ (v, k) }).collect::<Vec<(usize, usize)>>());
    let mut av: Vec<usize> = Vec::new();
    for (k, mut v) in am {
        let mut ret = 0usize;
        while v > 0 {
            let b = bh.pop();
            if let Some((bv, bk)) = b {
                if k == bk {
                    ret = bv;
                } else {
                    let smaller = v.min(bv);
                    for _ in 0..smaller {
                        //writeln!(io, "{} {} {} {}", k, v, bk, bv).unwrap();
                        av.push(bk);
                    }
                    v -= smaller;
                    if bv > smaller {
                        bh.push((bk, bv-smaller));
                    }
                }
            } else {
                writeln!(io, "No").unwrap();
                return
            }
        }
        if ret != 0 {
            bh.push((ret, k));
        }
    }
    writeln!(io, "Yes").unwrap();
    for (i, a) in av.iter().enumerate() {
        write!(io, "{}{}", a, if i == av.len()-1 {"\n"} else {" "}).unwrap();
    }
}