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::input;

#[allow(unused_imports)]
use proconio::marker::{Bytes, Chars, Usize1};
#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
use std::ops::*;

#[derive(Clone, Debug, Default)]
struct Solver {}

impl Solver {
    fn solve(&mut self) {
        input! {
            n: usize,
            a: [usize;n],
            mut b: [usize;n],
        }

        b.reverse();
        let mut conflict = None;
        for i in 0..n {
            if a[i] == b[i] {
                conflict = Some((i, a[i]));
                break;
            }
        }
        if let Some((p2, val)) = conflict {
            let level = |i| if a[i] == val { 1 } else { 0 } + if b[i] == val { 1 } else { 0 };
            let p1 = (0..p2).filter(|&i| level(i) == 0).count();
            let p3 = p2 + (p2..n).filter(|&i| level(i) == 2).count();
            let p4 = p3 + (p3..n).filter(|&i| level(i) == 1).count();

            if (p1 - 0) + (n - p4) < p3 - p2 {
                println!("No");
                return;
            }
            let p = min( p2 + (p1 - 0), p3 );
            &b[0..p].reverse();
            &b[p..n].reverse();
        }

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

//#[proconio::fastout]
fn main() {
    let mut s: Solver = Default::default();
    s.solve();
}