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;
#define ll long long
#define Mod 1000000007
#define Max 1000001
#define vt vector<long long>

void func(){
    ll n;
    cin>>n;
    ll a[n];
    ll b[n];
    unordered_map<ll, ll>ma, mb;
    for(int i=0; i<n; i++){
        
        cin>>a[i];
        ma[a[i]]++;
    }
    for(int i=0; i<n; i++){
        cin>>b[i];
        mb[b[i]]++;
    }
    vector<ll>v;
    ll i=0;
    ll j=n-1;
    
    while(v.size()<n){
        if(a[i]!=b[j] && mb[b[j]]!=0){
            v.push_back(b[j]);
            mb[b[j]]--;
            i++;
            j--;
        }
        else if(a[i]==b[j]){
            bool ok=false;
            for(int k=0; k<j; k++){
                if(a[i]!=b[k] && mb[b[k]]!=0){
                    v.push_back(b[k]);
                    mb[b[k]]--;
                    ok=true;
                    break;
                }
            }
            if(!ok){
                cout<<"No"<<endl;
                return;
            }
            i++;
        }
        else if(a[i]!=b[j] && mb[b[j]]==0){
            bool ok=false;
            for(int k=0; k<j; k++){
                if(a[i]!=b[k] && mb[b[k]]!=0){
                    v.push_back(b[k]);
                    mb[b[k]]--;
                    ok=true;
                    break;
                }
            }
            if(!ok){
                cout<<"No"<<endl;
                return;
            }
            i++;
        }
    }
    cout<<"Yes"<<endl;
    for(auto i: v){
        cout<<i<<" ";
    }
    cout<<endl;
}

int main(){
 
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
 

// 
 int t;
 cin >> t;
 for(ll i=1; i<=t; i++)
 {
    func();
//    cout<<endl;
//cout<<func()<<endl;
 
//if(func())
//    cout<<"Yes"<< endl;
//else
//    cout <<"No"<<endl; 
//cout<<"Case"<<" "<<"#"<<i<<":"<<" "<<func()<<endl;
 }
return 0;
}