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)

#![allow(unused_imports, non_snake_case)]
use proconio::{
    input, fastout,
    marker::{Chars, Bytes, Usize1}
};

#[fastout]
fn main() {
    input!{
        N: usize,
        A:[usize; N],
        mut B:[usize; N],
    }
    B.reverse();
    let mut f = vec![false; N];
    let mut idx = 0;
    for i in 0..N {
        if A[i] != B[i] { continue };
        while idx < N && (A[idx] == A[i] || B[idx] == B[i] || f[i]) {
            idx += 1;
        }
        if idx >= N {
            println!("No");
            return;
        }
        B.swap(i, idx);
        f[i] = true;
        f[idx] = true;
        idx += 1;
    }
    println!("Yes");
    for &x in &B {
        print!("{} ", x);
    }
    println!();
}