Problem: 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.

Examples

Input

6
1 1 1 2 2 3
1 1 1 2 2 3


Output

Yes
2 2 3 1 1 1


Input

3
1 1 2
1 1 3


Output

No


Input

4
1 1 2 3
1 2 3 3


Output

Yes
3 3 1 2

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class Main {

	// Shojin
	public static void main(String[] args) {
		try (Scanner sc = new Scanner(System.in)) {
			int N = sc.nextInt();
			int[] B = new int[N];
			int max = 0;
			int[] countArrA = new int[N + 1];
			int[] countArrB = new int[N + 1];
			for (int i = 0; i < N; i++) {
				int next = sc.nextInt();
				max = Math.max(max, next);
				countArrA[next]++;
			}
			for (int i = 0; i < N; i++) {
				int next = sc.nextInt();
				B[i] = next;
				max = Math.max(max, next);
				countArrB[next]++;
			}
			for (int i = 1; i <= max; i++) {
				if (countArrA[i] + countArrB[i] > N) {
					System.out.println("No");
					return;
				}
				countArrA[i] += countArrA[i - 1];
				countArrB[i] += countArrB[i - 1];
			}
			int move = 0;
			for (int i = 1; i <= max; i++) {
				move = Math.max(countArrA[i] - countArrB[i - 1], move);
			}
			System.out.println("Yes");
			for (int i = 0; i < N; i++) {
				int idx = (i + N - move) % N;
				System.out.print(B[idx]);
				System.out.print(' ');
			}
		}
	}
}