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: Time Limit Exceeded)

#![allow(
  non_snake_case,
  unused_variables,
  unused_assignments,
  unused_mut,
  unused_imports,
  unused_macros,
  dead_code
)]
//use proconio::fastout;
use proconio::{input, marker::*};
use std::cmp::*;
use std::collections::*;
//use std::collections::VecDeque;

macro_rules! debug {
  ($($a:expr),* $(,)*) => {
      #[cfg(debug_assertions)]
      eprintln!(concat!($("| ", stringify!($a), "={:?} "),*, "|"), $(&$a),*);
  };
}

fn main() {
  input! {
     N:usize,A:[usize;N],B:[usize;N]
  }
  let N: usize = N;
  let A: Vec<usize> = A;
  let B: Vec<usize> = B;
  let mut n = vec![0; N + 1];
  let mut d = 0;
  for i in 0..N {
    n[A[i]] += 1;
    n[B[i]] += 1;
    while A[i] == B[(i + d) % N] {
      d += 1;
      if d >= N {
        println!("No");
        return;
      }
    }
  }
  for i in 0..N {
    if n[i] > N {
      println!("No");
      return;
    }
  }

  for i in 0..N * N {
    while A[i % N] == B[(i + d) % N] {
      d += 1;
      if d >= N {
        println!("No");
        return;
      }
    }
  }
  for i in 0..N {
    if A[i] == B[(i + d) % N] {
      panic!("crash and burn");
      println!("No");
      return;
    }
  }
  println!("Yes");
  for i in 0..N {
    print!("{} ", B[(i + d) % N]);
  }
}