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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
//Optimisations
#pragma GCC target ("avx2")
#pragma GCC optimization ("unroll-loops")
#pragma GCC optimize("O2")
//shortcuts for functions
#define pb push_back
#define mp make_pair
#define ff first 
#define ss second
#define all(v) v.begin(),v.end()
#define prec(n) fixed<<setprecision(n)
#define n_l '\n'
// make it python 
#define gcd __gcd
#define append push_back
#define str to_string
// utility functions shortcuts
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define sswap(a,b) {a=a^b;b=a^b;a=a^b;}
#define swap(a,b) {auto temp=a; a=b; b=temp;}
#define init(dp) memset(dp,-1,sizeof(dp));
#define set0(dp) memset(dp,0,sizeof(dp));
#define bits(x) __builtin_popcount(x)
#define SORT(v) sort(all(v))
#define endl "\n"
#define forr(i,n) for(ll i=0;i<n;i++)
// declaration shortcuts
typedef long long int ll;
#define int ll
// Constants
constexpr int dx[] = {-1, 0, 1, 0, 1, 1, -1, -1};
constexpr int dy[] = {0, -1, 0, 1, 1, -1, 1, -1};
constexpr ll INF = 1999999999999999997; 
constexpr int inf= INT_MAX;
constexpr int MAXSIZE = int(1e6)+5;
constexpr auto PI  = 3.14159265358979323846L;
constexpr auto oo  = numeric_limits<int>::max() / 2 - 2;
constexpr auto eps = 1e-6;
constexpr auto mod = 1000000007;
constexpr auto MOD = 1000000007;
constexpr auto MOD9 = 1000000009;
constexpr auto maxn = 100006;
//void IOfile(){
//freopen(file_name, reade_mode, stdin);
//freopen(file_name, write_mode, stdout);
//}
void fastio(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
}
int32_t main(){
	fastio();
	int n;
	cin >> n;
	int a[n];
	forr(i,n) cin >> a[i];
	map<int,int>f;
	for(int i = 0; i < n; i++){
		int x;
		cin >> x;
		f[x]++;
	}
	int ans[n];
	map<int, int>::iterator  j;
	
	for(int i = 0; i < n; i++){
		bool ok = false;
		
		
		for( j= --f.end(); j != f.begin(); j--){
			if(j->first != a[i]){
				ok = true;
				ans[i] = j -> first;
				//cout << i << " " << ans[i] <<  " " << j->first << "me ran\n";
				 f[j->first]--;
				 if(f[j->first] == 0){
				 	f.erase(j->first);
				 }
				 break;
			}
		}
		//cout << i << " " << ok << "i ran\n";
		if(j == f.begin() and !ok){
			if(f.begin()->first == a[i]){
				cout << "No";
				return 0;
			}else{
				ok = true;
				ans[i] = f.begin()->first;
				f.begin()->second--;
			}
			
		}
		if(!ok){
			cout << "No";
			return 0;
		}
		//cout << i << " " << ans[i] << "\n";
	}
	cout << "Yes\n";
	for(auto i:ans) cout << i << " ";
}