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(
  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 mut B: Vec<usize> = B;
  // 各数値iについて、合計個数nとAB内での開始位置と終了位置(開区間)をAi,Aj,Bi,Bjで並べる。
  let mut v = HashMap::new();
  let mut ai = 0;
  let mut bi = 0;
  let mut diffmax = 0;
  while ai < N {
    // 開始位置をそろえる
    let a = A[ai];
    while bi < N && B[bi] < a {
      bi += 1;
    }
    // 数える
    let mut aj = ai;
    while aj < N && A[aj] == a {
      aj += 1;
    }
    let mut bj = bi;
    while bj < N && B[bj] == a {
      bj += 1;
    }
    let n = aj - ai + bj - bi;
    v.insert(a, (n, ai, aj, bi, bj));
    if aj > bi {
      diffmax = max(diffmax, aj - bi);
    }
    if n > N {
      println!("No");
      return;
    }
    ai = aj;
    bi = bj;
  }
  println!("Yes");
  for i in diffmax..N {
    print!("{} ", B[i]);
  }
  for i in 0..diffmax {
    print!("{} ", B[i]);
  }
  debug!(v, diffmax);
}