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: Accepted)

use proconio::*;
use std::cmp::*;
use std::collections::*;
use std::ops::Bound::*;

#[fastout]
fn run() {
    input! {
        n: usize,
        a: [usize; n],
        b: [usize; n],
    }
    let mut range = vec![(n, 0); n + 1];
    for (i, a) in a.iter().enumerate() {
        let po = &mut range[*a];
        po.0 = min(po.0, i);
        po.1 = i;
    }
    let mut set = (0..n).collect::<BTreeSet<_>>();
    for (i, &b) in b.iter().enumerate() {
        let (x, y) = range[b];
        if x == n {
            continue;
        }
        if x >= i {
            let mut del = vec![];
            for &v in set.range((Included(&(x - i)), Included(&(y - i)))) {
                del.push(v);
            }
            for v in del {
                set.remove(&v);
            }
        } else {
            let mut del = vec![];
            for &v in set.range((Included(&(n + x - i)), Included(&(n + y - i)))) {
                del.push(v);
            }
            if y >= i {
                for &v in set.range((Included(&0), Included(&(y - i)))) {
                    del.push(v);
                }
            }
            for v in del {
                set.remove(&v);
            }
        }
    }
    if let Some(&k) = set.iter().next() {
        println!("Yes");
        for (i, b) in b.iter().cycle().skip(n - k).take(n).enumerate() {
            if i > 0 {
                print!(" ");
            }
            print!("{}", b);
        }
        println!();
    } else {
        println!("No");
    }
}

fn main() {
    run();
}