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:
//package src.com.jetman.atcoder;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

/**
 * @description:
 * @author: Jetman
 * @create: 2020-08-29 14:39
 **/
public class Main {

    Scanner sc = new Scanner(System.in);

    long MOD = 1000000000 + 7;


    private void slove() {

        int n = sc.nextInt();

        int[] a = new int[n];
        int[] b = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        for (int i = 0; i < n; i++) {
            b[n-i-1] = sc.nextInt();
        }

        int c = -1;
        for (int i = 0; i < n; i++) {
            if (a[i] == b[i]) {
                c = a[i];
                break;
            }
        }
        int l = 0;
        for (int i = 0; i < n; i++) {
            if (a[i] == c && b[i] == c) {
                l = i;
                break;
            }
        }
        int r = -1;
        for (int i = n-1; i >=0; i--) {
            if (a[i] == c && b[i] == c) {
                r = i;
                break;
            }
        }

        for (int i = 0; i < n; i++) {
            if (c != -1 && a[i] != c && b[i] != c && l <= r) {
                int tmp = b[i];
                b[i] = b[l];
                b[l] = tmp;
                l++;
            }
        }
        if (l <= r) {
            System.out.println("No");
        } else {
            System.out.println("Yes");
            for (int i = 0; i < n; i++) {
                System.out.print(b[i] + " ");
            }
        }








    }

    public static void main(String[] args) {
        new Main().slove();
    }
}