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)

use proconio::input;
use std::cmp::min;
use std::collections::HashMap;

fn main() {
    input! {
        n : usize,
        mut a : [usize;n],
        mut b : [usize;n]
    }
    b.reverse();
    let mut idx = n;
    let mut map: HashMap<usize, usize> = HashMap::new();
    let mut cnt = 0;
    let mut num = 210000;
    for i in 0..n {
        if a[i] == b[i] {
            idx = min(i, idx);
            cnt += 1;
            num = a[i];
        }
        *map.entry(a[i]).or_insert(0) += 1;
        *map.entry(b[i]).or_insert(0) += 1;
    }
    for (_k, v) in map {
        if v > n {
            println!("No");
            return;
        }
    }
    println!("Yes");
    if cnt == 0 {
        for i in b {
            print!("{} ", i);
        }
        return;
    }
    for i in 0..n {
        if b[i] != num && b[idx] == num
        {
            b.swap(i,idx);
            idx+=1;
        }else
        {
            break;
        }
    }
    let cidx = idx;
    for i in (cidx..n).rev() {
        if b[i] != num && b[idx] == num
        {
            b.swap(i,idx);
            idx+=1;
        }
        if i <= idx {
            break;
        }
    }
    for i in b {
        print!("{} ",i);
    }
}