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)


import java.util.ArrayDeque;
import java.util.Queue;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = Integer.parseInt(scan.next());
        int[] a = new int[n];
        Queue<Integer> b = new ArrayDeque<Integer>();
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(scan.next());
        }
        for (int i = 0; i < n; i++) {
            b.add(Integer.parseInt(scan.next()));
        }
        scan.close();

        StringBuilder sb = new StringBuilder();
        int cursor = 0;
        int loopCnt = 0;
        while (cursor < n) {
            int now = b.poll();
            if (a[cursor] != now) {
                sb.append(now);
                sb.append(" ");
                cursor++;
                loopCnt = 0;
            } else {
                b.add(now);
                loopCnt++;
            }

            if (loopCnt > n - cursor) {
                System.out.println("No");
                return;
            }
        }

        System.out.println("Yes");
        System.out.println(sb.toString());
    }
}