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>
#define int long long
#define fi first
#define se second
#define vll vector<int>
#define pii pair<int,int>
#define pb push_back
#define sz(v) (int)(v).size()
#define inf 1e18
#define md 1000000007
#define all(v) (v).begin(),(v).end()
#define rep(i,a,b) for(int i=a;i<b;++i)
using namespace std;

#define M 100010


int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
   	
   	int n;cin>>n;
   	vector<pii> a(n);
   	int b[n],c[n];
   	map<int,int>mp;
   	vll id[n+1];
   	rep(i,0,n)cin>>c[i],mp[c[i]]++,id[c[i]].pb(i);
   	set<pii>s;
   	for(auto p:mp)s.insert({p.se,p.fi});
   	int i=n-1;
   	while(i>=0){
   		auto it=s.begin();
   		pii p=*it;
   		s.erase(it);
   		int x=p.se;
   		rep(k,0,p.fi)
   			a[i].fi=x,a[i].se=id[x].back(),id[x].pop_back(),i--;
   	}
   	
   	// rep(i,0,n)cout<<a[i].fi<<" ";cout<<"\n";
   	mp.clear(),s.clear();
   	rep(i,0,n)cin>>b[i],mp[b[i]]++;
   	for(auto p:mp)s.insert({p.se,-p.fi});
   rep(i,0,n)b[i]=-1;
   	rep(i,0,n){
   		int x=a[i].fi,j=a[i].se;
   		auto it=s.end();it--;
   		pii p=*it;p.se=-p.se;
   		if(p.se!=x){
   			s.erase(it);
   			if(p.fi>1)s.insert({p.fi-1,-p.se});
   			b[j]=p.se;
   		}
   		else{
   			if(s.size()==1){
   				cout<<"No\n";
   				rep(i,0,n)cout<<b[i]<<" ";
   				return 0;
   			}
   			it--,p=*it,s.erase(it),p.se=-p.se;
   			if(p.fi>1)s.insert({p.fi-1,-p.se});
   			b[j]=p.se;
   		}
   	}
   	rep(i,0,n){
   		if(b[i]==c[i]){
   			cout<<"No";
   			return 0;
   		}
   	}
   	cout<<"Yes\n";
   	rep(i,0,n)cout<<b[i]<<" ";
   	
   	
    return 0;
}