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>
#include<string.h>
#define ll long long 
#include<vector>
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
const int mod=1e9+7; 
int main(){
	IOS;
	int n;
	cin>>n;
	int a_map[200005];
	int b_map[200005];
	memset(a_map,0,sizeof(a_map));
	memset(b_map,0,sizeof(b_map));
	//vector<int> b_map(n+1,0);
	for(int i=0;i<n;i++)
	{
		int a;
		cin>>a;
		a_map[a]++;
	}
	
	for(int i=0;i<n;i++){
		int b;
		cin>>b;
		b_map[b]++;
	}
	bool flag=false;
	for(int i=0;i<n+1;i++){
		int can_place=0;
		if(a_map[i]){
			can_place=n-b_map[i];
			if(can_place<a_map[i])
			{
				flag=true;
				break;
			}
		}
	}
	if(flag)
	cout<<"No";
	else{
		int cnt=0;
		vector<int> ans(n);
		cout<<"Yes\n";
		for(int i=n;i>=0;i--){
			if(b_map[i]!=0){
				cnt+=b_map[i];
				for(int j=0;j<b_map[i];j++){
					cout<<i<<" ";
				}
				if(cnt==n)
				break;
			}
		}
	}
	
}