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: Accepted)

import java.io.*;
import java.util.*;

class Main
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int n=Integer.parseInt(bu.readLine());
        int a[]=new int[n],b[]=new int[n],i;
        String s[]=bu.readLine().split(" ");
        for(i=0;i<n;i++)
        a[i]=Integer.parseInt(s[i]);
        s=bu.readLine().split(" ");
        for(i=0;i<n;i++)
        b[i]=Integer.parseInt(s[i]);

        Queue<Integer> q=new LinkedList<>();
        for(i=0;i<n;i++)
        {
            if(a[i]==b[i])
            {
                if(q.isEmpty()) q.add(i);
                else
                {
                    if(a[q.peek()]!=b[i])
                    {
                        int t=q.poll();
                        b[t]=b[i];
                        b[i]=a[t];
                    }
                    else q.add(i);
                }
            }
        }

        int sz=0;
        while(!q.isEmpty() && q.size()!=sz)
        {
            sz=q.size();
            for(i=0;i<n;i++)
            if(a[i]!=b[i] && !q.isEmpty() && a[i]!=a[q.peek()] && b[i]!=a[q.peek()])
            {
                int t=q.poll();
                b[t]=b[i];
                b[i]=a[t];
            }
        }
        if(!q.isEmpty()) sb.append("No");
        else
        {
            sb.append("Yes\n");
            for(i=0;i<n;i++)
            sb.append(b[i]+" ");
        }
        System.out.print(sb);
    }
}