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.Arrays;
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（貪欲法よる解法）
 */
public class Main {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt();
			int[] a = IntStream.range(0, n).map(i -> scanner.nextInt()).toArray();
			int[] b = IntStream.range(0, n).map(i -> scanner.nextInt()).toArray();
			int j = 0, bef = -1;
			for (int i = 0; i < n; i++) {
				if (bef != a[i]) {
					j = 0;
				}
				if (a[i] == b[i]) {
					for (; j < n; j++) {
						if ((a[j] != a[i]) && (b[j] != a[i])) {
							swap(b, i, j);
							break;
						}
					}
					if (a[i] == b[i]) {
						System.out.println("No");
						return;
					}
				}
				bef = a[i];
			}
			System.out.println("Yes");
			System.out.println(Arrays.stream(b).mapToObj(String::valueOf).collect(Collectors.joining(" ")));
		}
	}

	/**
	 * 配列のi番目のものとj番目のものを交換する
	 * 
	 * @param array 配列
	 * @param i
	 * @param j
	 */
	private static void swap(int[] array, int i, int j) {
		int temp = array[i];
		array[i] = array[j];
		array[j] = temp;
	}
}