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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

//created by Whiplash99
import java.io.*;
import java.sql.Array;
import java.util.*;
public class Main
{
    public static void main(String[] args) throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

        int i,N;

        N=Integer.parseInt(br.readLine().trim());
        String[] s=br.readLine().trim().split(" ");
        int a[]=new int[N];
        for(i=0;i<N;i++) a[i]=Integer.parseInt(s[i]);

        s=br.readLine().trim().split(" ");
        int b[]=new int[N];
        for(i=0;i<N;i++) b[i]=Integer.parseInt(s[i]);

        HashMap<Integer,Integer> map=new HashMap<>();
        ArrayDeque<Integer> Q=new ArrayDeque<>();
        for(i=0;i<N;i++)
        {
            Q.add(b[i]);
            map.put(b[i],map.getOrDefault(b[i],0)+1);
        }
        boolean flag=true;

        for(i=0;i<N;i++)
        {
            if(a[i]==Q.peek())
            {
                if(map.size()==1)
                {
                    flag=false;
                    break;
                }
                while (Q.peek()==a[i]) Q.add(Q.poll());
            }
            a[i]=Q.poll();
            if(map.get(a[i])==1) map.remove(a[i]);
            else map.put(a[i],map.get(a[i])-1);
        }
        if(!flag) System.out.println("No");
        else
        {
            System.out.println("Yes");
            StringBuilder sb=new StringBuilder();
            for(i=0;i<N;i++) sb.append(a[i]).append(" ");
            System.out.println(sb);
        }
    }
}