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 itertools::Itertools;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;

fn main() {
    input! {
        n: usize,
        a: [i64; n],
        b: [i64; n],
    }
    use std::collections::{BTreeMap, BinaryHeap};
    let mut cnta = BTreeMap::new();
    let mut cntb = BTreeMap::new();
    for i in 0..n {
        cnta.entry(a[i]).or_insert((0, i)).0 += 1;
        cntb.entry(b[i]).or_insert((0, i)).0 += 1;
    }
    let cnta = cnta
        .iter()
        .map(|(a, (c, i))| ((*c, *a), *i))
        .collect::<BTreeMap<_, _>>();
    let mut cntb = cntb
        .iter()
        .map(|(a, (c, _))| (*c, *a))
        .collect::<BinaryHeap<_>>();

    let mut res = vec![0; n];
    for ((c, a), i) in cnta.iter().rev() {
        let mut tmp = None;
        for j in 0..*c {
            let x = cntb.pop();
            if x == None {
                println!("No");
                return;
            }
            let mut x = x.unwrap();
            if x.1 == *a {
                tmp = Some(x);
                if let Some(y) = cntb.pop() {
                    x = y;
                } else {
                    println!("No");
                    return;
                }
            }
            res[i + j] = x.1;
            if x.0 > 1 {
                cntb.push((x.0 - 1, x.1));
            }
        }
        if let Some(x) = tmp {
            cntb.push(x);
        }
    }

    println!("Yes");
    for i in 0..n {
        print!("{} ", res[i]);
    }
}