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: Runtime Error)

#![allow(unused_imports)]

// language: Rust(1.42.0)
// check available crates on AtCoder at "https://atcoder.jp/contests/language-test-202001"
// My Library Repositry is at "https://github.com/Loptall/sfcpl"

use itertools::*;
use itertools_num::*;
use lazy_static::lazy_static;
use maplit::{btreemap, btreeset, hashmap, hashset};
use num_bigint::{BigInt, BigUint};
use num_complex::Complex;
use num_integer::{binomial, gcd, lcm, multinomial, Integer};
use num_rational::Rational;
use num_traits::{
    clamp, one, pow, zero, Num, NumAssignOps, NumOps, One, Pow, Signed, Unsigned, Zero,
};
use permutohedron::Heap;
use proconio::{
    derive_readable, fastout, input, is_stdin_empty,
    marker::{Bytes, Chars, Isize1, Usize1},
};
use rand::random;
use std::cmp::{
    max, min, Ordering,
    Ordering::{Equal, Greater, Less},
    Reverse,
};
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
use std::convert::TryInto;
use std::fmt;
use std::mem::swap;
use std::num::{NonZeroU32, ParseIntError};
use std::ops::{
    Add, AddAssign, BitAnd, BitAndAssign, BitOr, BitOrAssign, BitXor, BitXorAssign, Div, DivAssign,
    Index, IndexMut, Mul, MulAssign, Neg, Rem, RemAssign, Shl, ShlAssign, Shr, ShrAssign, Sub,
    SubAssign,
};
use std::process::exit;
use std::{f32, f64, i128, i16, i32, i64, i8, isize, u128, u16, u32, u64, u8, usize};

pub use visualize::*;
pub mod visualize {
    use itertools::*;
    pub trait Visualize {
        fn visualize(&self, split: &str);
        fn continuous(&self) {
            self.visualize("");
        }
        fn spaces(&self) {
            self.visualize(" ");
        }
        fn lines(&self) {
            self.visualize("\n");
        }
    }
    macro_rules ! impl_vis_for_sized {($ ($ t : ty ) ,+ ) => {$ (impl Visualize for $ t {fn visualize (& self , _split : & str ) {print ! ("{}" , self ) ; } } ) + } ; }
    impl_vis_for_sized! {usize , u8 , u16 , u32 , u64 , u128 , isize , i8 , i16 , i32 , i64 , i128 , String , & str , char }
    impl<T: std::fmt::Display> Visualize for [T] {
        fn visualize(&self, split: &str) {
            print!("{}", self.iter().join(split));
        }
    }
    #[macro_export]
    macro_rules ! vis {() => {println ! () ; } ; ($ last : expr ; ) => {$ last . lines () ; vis ! () } ; ($ last : expr => ) => {$ last . continuous () ; vis ! () ; } ; ($ last : expr $ (, ) ? ) => {$ last . spaces () ; vis ! () ; } ; ($ first : expr ; $ ($ rest : tt ) * ) => {$ first . lines () ; println ! () ; vis ! ($ ($ rest ) * ) ; } ; ($ first : expr => $ ($ rest : tt ) * ) => {$ first . continuous () ; vis ! ($ ($ rest ) * ) ; } ; ($ first : expr , $ ($ rest : tt ) * ) => {$ first . spaces () ; print ! (" " ) ; vis ! ($ ($ rest ) * ) ; } ; }
}

pub use consts::*;
pub mod consts {
    pub const MOD10E9_7: usize = 1000000007; // 10 ^ 9 + 7
    pub const MOD99_: usize = 998244353;
    pub const MAX: usize = std::usize::MAX; // = 2 ^ 64 - 1 = 18446744073709551615 ≈ 1.8 * 10 ^ 19
    pub const INF: usize = 2000000000000000000; // MAX / 9 < 2 * 10e18 < MAX / 10
    pub const FNI: i64 = -2000000000000000000; // == -(INF as i64)
    pub const PI: f64 = std::f64::consts::PI; // 3.141592653589793 -- 10 ^ -15
    pub const ASCII_A_LARGE: u8 = 65;
    pub const ASCII_A_SMALL: u8 = 97;
    pub const ASCII_0: u8 = 48;
    pub const ADJ4: &[(isize, isize); 4] = &[(1, 0), (0, 1), (-1, 0), (0, -1)];
    pub const ADJ8: &[(isize, isize); 8] = &[
        (1, 0),
        (1, 1),
        (0, 1),
        (-1, 1),
        (-1, 0),
        (-1, -1),
        (0, -1),
        (1, -1),
    ];
}

// snippet here


// #[fastout]
fn main() {
    input! {
        n: usize,
        a: [usize; n],
        b: [usize; n]
    }

    let mut count = vec![0usize; n+1];
    for i in a.iter() {
        count[*i] += 1;
    }
    for i in b.iter() {
        count[*i] += 1;
    }

    if count.into_iter().any(|x| x > n) {
        vis!("No");
        return;
    } else {
        vis!("Yes");
    }

    let mut c = 0usize;
    let mut d = 0usize;
    for &e in unique_count(&b).keys() {
        let bef = b.iter().enumerate().find(|x| *x.1 == e).unwrap().0;
        let aft = a.iter().enumerate().skip_while(|x| *x.1 != e).find(|x| *x.1 != e).unwrap_or((bef, &0)).0;
        d = max(d, aft - bef);
    }

    let ans = {
        let mut b = b;
        b.rotate_right(d);
        b
    };
    vis!(ans);

}

/// 配列に含まれる要素(T: Ord + Copy)別にその個数を数えて
/// `BTreeMap<T, usize>`の形にして返す
pub fn unique_count<T: Ord + Copy>(v: &[T]) -> BTreeMap<T, usize> {
    let mut map = BTreeMap::new();
    for e in v {
        let h = map.entry(*e).or_insert(0);
        *h += 1;
    }
    map
}