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>

/*
#pragma GCC optimize("-Ofast")
//#pragma GCC optimize("trapv")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.2,popcnt,abm,mmx,avx2,tune=native")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-funroll-loops")*/

#define I inline void 
#define S struct 
#define vi vector<int> 
#define vii vector<pair<int,int>>
#define pii pair<int,int>
#define pll pair<ll,ll>

using namespace std ; 
using ll = long long ; 
using ld = long double ; 

const int N = 2e5 + 7 , mod = 1e9 + 7 ; 
const int inf = N ; 
// How interesting!

int n ;

vector<int> a(N , 0) , b(N , 0); 
int occ[N] , occ2[N] ; 
int mx1 , mx2;  

int ans[N] ; 

I check(){
        for(int i = 0 ;i < n ;i++){
                if(a[i] == b[i]){
                        return ; 
                }
        }
        cout<<"Yes\n"; 
        for(int i = 0 ;i < n;i++)
                cout<< b[i] <<" " ;
        exit(0) ;  
}

I shift(vector<int> &aux , int x){
        vector<int> tmp(n , 0) ; 
        for(int i = 0 ;i < n;i++){
                tmp[(i + x + n) %n] = aux[i] ; 
        }
        for(int i = 0 ;i < n;i++){
                aux[i] = tmp[i] ; 
        }
}

I make(int x){
        shift(b , x) ; 
        check() ; 
        shift(b , -x) ; 
}

int main(){
        ios_base::sync_with_stdio(0) ; 
        cin.tie(0) ;
        //freopen("in.in" , "r" , stdin) ;
        cin >> n; 

        for(int i = 0 ;i < n;i ++ ){
                cin >> a[i] ; 
                mx1 = max(mx1 , ++occ[a[i]]) ; 
        }

        for(int i = 0 ;i < n;i++){
                cin >> b[i] ; 
                mx2 = max(mx2 , ++occ2[b[i]]) ; 
        }
        make(mx1) ; 
        make(mx2) ; 
        make(n - mx1) ; 
        make(n - mx2) ; 
        cout<<"No" ; 
        return 0 ; 
}