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 C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#line 2 "/Users/kaage/Desktop/ProgrammingWorkspace/library/other/template.hpp"
#define _CRT_SECURE_NO_WARNINGS
#pragma target("avx2")
#pragma optimize("O3")
#pragma optimize("unroll-loops")
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <ctime>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <list>
#include <map>
#include <memory>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <string.h>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define all(V) V.begin(),V.end()
typedef unsigned int uint;
typedef long long lint;
typedef unsigned long long ulint;
typedef std::pair<int, int> P;
typedef std::pair<lint, lint> LP;
constexpr int INF = INT_MAX/2;
constexpr lint LINF = LLONG_MAX/2;
constexpr double eps = DBL_EPSILON;
constexpr double PI=3.141592653589793238462643383279;
template<class T>
class prique :public std::priority_queue<T, std::vector<T>, std::greater<T>> {};
template <class T, class U>
inline bool chmax(T& lhs, const U& rhs) {
	if (lhs < rhs) {
		lhs = rhs;
		return 1;
	}
	return 0;
}
template <class T, class U>
inline bool chmin(T& lhs, const U& rhs) {
	if (lhs > rhs) {
		lhs = rhs;
		return 1;
	}
	return 0;
}
inline lint gcd(lint a, lint b) {
	while (b) {
		lint c = a;
		a = b; b = c % b;
	}
	return a;
}
inline lint lcm(lint a, lint b) {
	return a / gcd(a, b) * b;
}
bool isprime(lint n) {
	if (n == 1)return false;
	for (int i = 2; i * i <= n; i++) {
		if (n % i == 0)return false;
	}
	return true;
}
template<typename T>
T mypow(T a, lint b) {
	T res(1);
	while(b){
		if(b&1)res*=a;
		a*=a;
		b>>=1;
	}
	return res;
}
lint modpow(lint a, lint b, lint m) {
	lint res(1);
	while(b){
		if(b&1){
			res*=a;res%=m;
		}
		a*=a;a%=m;
		b>>=1;
	}
	return res;
}
template<typename T>
void printArray(std::vector<T>& vec) {
	rep(i, vec.size()){
		std::cout << vec[i];
		std::cout<<(i==(int)vec.size()-1?"\n":" ");
	}
}
template<typename T>
void printArray(T l, T r) {
	T rprev = std::prev(r);
	for (T i = l; i != rprev; i++) {
		std::cout << *i << " ";
	}
	std::cout << *rprev << std::endl;
}
LP extGcd(lint a,lint b) {
	if(b==0)return {1,0};
	LP s=extGcd(b,a%b);
	std::swap(s.first,s.second);
	s.second-=a/b*s.first;
	return s;
}
LP ChineseRem(const lint& b1,const lint& m1,const lint& b2,const lint& m2) {
	lint p=extGcd(m1,m2).first;
	lint tmp=(b2-b1)*p%m2;
	lint r=(b1+m1*tmp+m1*m2)%(m1*m2);
	return std::make_pair(r,m1*m2);
}
template<typename F>
inline constexpr decltype(auto) lambda_fix(F&& f){
	return [f=std::forward<F>(f)](auto&&... args){
		return f(f,std::forward<decltype(args)>(args)...);
	};
}
#line 2 "main.cpp"
int N,A[200010],B[200010],C[200010],ans[200010];
std::set<int> st;
int main(){
	std::cin>>N;
	rep(i,N){
		std::cin>>A[i];
		C[A[i]]++;
	}
	rep(i,N){
		std::cin>>B[i];
		C[B[i]]++;
	}
	REP(i,N){
		if(C[i]>N){
			puts("No");
			return 0;
		}
	}
	rep(i,N)C[A[i]]--;
	std::sort(B,B+N,[&](const int& lhs,const int& rhs){
		return C[lhs]>C[rhs];
	});
	rep(i,N)st.insert(i);
	int now=0;
	while(now<N){
		for(auto ite=st.begin();ite!=st.end();){
			if(A[*ite]!=B[now]){
				ans[*ite]=B[now++];
				ite=st.erase(ite);
				if(now==N)break;
			}
			else ite++;
		}
	}
	puts("Yes");
	printArray(ans,ans+N);
	return 0;
}