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 1.8.0) (judge verdict: Wrong Answer)

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

@SuppressWarnings("unchecked")
public class Main {
    private static final long MOD = (int)1e9 + 7;

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] a = new int[n];
        int[] b = new int[n];
        for (int i = 0; i < n; ++i) a[i] = in.nextInt();
        for (int i = 0; i < n; ++i) b[i] = in.nextInt();

        int[] c = new int[n];
        int[] vis = new int[n];
        Arrays.fill(c, -1);
        int j = 0;
        for (int i = 0; i < n; ++i) {
            while (true) {
                vis[j]++;
                if (vis[j] > 200) {
                    System.out.println("No");
                    return;
                }
                if (a[j] != b[i] && c[j] == -1) {
                    c[j] = b[i];
                    j = (j + 1) % n;
                    break;
                }
                j = (j + 1) % n;
            }
        }
        // for (int i = 0; i < n; ++i) {
        //     System.out.print(c[i] + " ");
        // }
        for (int i = 0; i < n; ++i) {
            if (c[i] == a[i]) {
                System.out.println("No");
                return;
            }
        }
        System.out.println("Yes");
        for (int i = 0; i < n; ++i) {
            System.out.print(c[i] + " ");
        }
        System.out.println();
    }
}