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)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::ops::Bound::*;
use itertools::Itertools;
use num_traits::clamp;
use ordered_float::OrderedFloat;
use proconio::{input, marker::*, fastout};
use superslice::*;

#[fastout]
fn main() {
    input! {
        n: usize,
        a: [usize; n],
        mut b: [usize; n],
    }
    b.reverse();
    let mut pos1 = n;
    let mut pos2 = n;
    let mut found = false;
    for i in 0..n {
        if a[i] == b[i] {
            if !found {
                pos1 = i;
                pos2 = i;
                found = true;
            } else {
                pos2 = i;
            }
        }
    }
    if !found {
        println!("Yes");
        for b in b {
            print!("{} ", b)
        }
        return;
    }
    let mut p = pos1;
    let lower = pos1;
    let upper = pos2 + 1;
    for i in 0..lower {
        if p > pos2 {
            break;
        }
        if b[i] == a[pos1] {
            continue;
        }
        b.swap(i, p);
        p += 1;
    }
    for i in upper..n {
        if p > pos2 {
            break;
        }
        if b[i] == a[pos1] {
            continue;
        }
        b.swap(i, p);
        p += 1;
    }
    for i in 0..n {
        if a[i] == b[i] {
            println!("No");
            return;
        }
    }
    println!("Yes");
    for b in b {
        print!("{} ", b);
    }

}