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.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

/**
 * 解説通りに実装したソースコード
 */
public class Main {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt();
			Map<Integer, Integer> aMap = new HashMap<>(), bMap = new HashMap<>();
			IntStream.range(0, n).forEach(i -> {
				int c = scanner.nextInt();
				aMap.put(c, aMap.getOrDefault(c, 0) + 1);
			});
			int[] b = IntStream.range(0, n).map(i -> {
				int c = scanner.nextInt();
				bMap.put(c, bMap.getOrDefault(c, 0) + 1);
				return c;
			}).toArray();
			// 可能かどうかを判定
			for (Integer key : aMap.keySet()) {
				if (aMap.get(key) + bMap.getOrDefault(key, 0) > n) {
					System.out.println("No");
					return;
				}
			}
			// aに出現するi以下の値の個数をc[i]、bに出現するi以下の値の個数をd[i]
			int[] c = new int[n + 1], d = new int[n + 1];
			Arrays.fill(c, 0);
			Arrays.fill(d, 0);
			aMap.entrySet().stream().forEach(entry -> c[entry.getKey()] += entry.getValue());
			bMap.entrySet().stream().forEach(entry -> d[entry.getKey()] += entry.getValue());
			IntStream.rangeClosed(1, n).forEach(i -> {
				c[i] += c[i - 1];
				d[i] += d[i - 1];
			});
			// bをc[i]−d[i−1] の最大値の分だけ右にずらす
			int max = IntStream.rangeClosed(1, n).map(i -> c[i] - d[i - 1]).max().getAsInt();
			int[] result = IntStream.range(0, n).map(i -> b[(max + i) % n]).toArray();
			System.out.println("Yes");
			System.out.println(Arrays.stream(result).mapToObj(String::valueOf).collect(Collectors.joining(" ")));
		}
	}
}