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: Time Limit Exceeded)

import java.util.*;
public class Main{
    
    static final int MOD = (int)1e9 + 7;
    
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        
        int n = Integer.parseInt(sc.next());
        int[] a = new int[n];
        int[] b = new int[n];
        HashMap<Integer, Integer> ha = new HashMap<>();
        HashMap<Integer, Integer> hb = new HashMap<>();
        for(int i=0; i<n; i++){
            a[i] = Integer.parseInt(sc.next());
            ha.put(a[i], ha.getOrDefault(a[i], 0)+1);
        }
        for(int i=0; i<n; i++){
            b[i] = Integer.parseInt(sc.next());
            hb.put(b[i], hb.getOrDefault(b[i], 0)+1);
        }
        
        Set<Integer> hbs = hb.keySet();
        for(int k : hbs){
            int tmpa = ha.getOrDefault(k, 0);
            int tmpb = hb.getOrDefault(k, 0);
            if(tmpa + tmpb > n){
                System.out.println("No");
                return;
            }
        }
        
        ArrayDeque<Integer> q = new ArrayDeque<>();
        for(int i=0; i<n; i++){
            q.addLast(b[n-i-1]);
        }
        
        System.out.println("Yes");
        for(int i=0; i<n; i++){
            int now = q.pollFirst();
            while(a[i] == now){
                q.addLast(now);
                now = q.pollFirst();
            }
            System.out.print(now);
            if(i!=n-1){
                System.out.print(" ");
            }
        }
        
    }
    
}