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, non_snake_case)]
use proconio::{
    input, fastout,
    marker::{Chars, Bytes, Usize1}
};

#[fastout]
fn main() {
    input!{
        N: usize,
        A:[usize; N],
        B:[usize; N],
    }
    let m = 200_001;
    let mut f = vec![0; m];
    let mut g = vec![0; m];
    for &x in &A {
        f[x] += 1;
    }
    for &x in &B {
        g[x] += 1;
    }
    let mut ans = 0;
    for i in 1..m {
        if f[i] + g[i] > N {
            println!("No");
            return
        }
        f[i] += f[i-1];
        g[i] += g[i-1];
        if f[i] > g[i-1] && ans < f[i] - g[i-1] {
            ans = f[i] - g[i-1];
        }
    }
    println!("Yes");
    let mut B = B;
    let mut h = B.split_off(ans);
    h.append(&mut B);
    println!("{}", h.iter().map(|x| x.to_string()).collect::<Vec<_>>().join(" "));
}