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

public class Main {
    public static void main(String[] args) throws Exception {
        BufferedReader buffer = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(buffer.readLine());
        String [] inp = buffer.readLine().split(" ");
        Integer []arr = new Integer[n];
        int [] arr2 = new int[n];
        for (int pos = 0; pos < n; pos++) {
            arr[pos] = Integer.parseInt(inp[pos]);
        }
        inp = buffer.readLine().split(" ");
        for (int pos = 0; pos < n; pos++) {
            arr2[pos] = Integer.parseInt(inp[pos]);
        }
        Queue<Integer>balancedPos = new LinkedList<>();
        for (int pos = 0; pos < n; pos++) {
            if (arr[pos] == arr2[pos]){
                if (!balancedPos.isEmpty() && arr[balancedPos.peek()]!=arr2[pos]){
                    int i = balancedPos.poll();
                    arr2[i] = arr2[pos]^arr2[i]^(arr2[pos] = arr2[i]);
                }
                else
                    balancedPos.add(pos);
            }
        }
        int size = 0;
        while (!balancedPos.isEmpty() && balancedPos.size()!=size){
            size = balancedPos.size();
            for (int pos = 0; pos < n; pos++) {
                if (arr[pos]!=arr2[pos] && !balancedPos.isEmpty() && arr[balancedPos.peek()]!=arr2[pos] && arr2[balancedPos.peek()]!=arr[pos]){
                    int i = balancedPos.poll();
                    arr2[i] = arr2[i]^arr2[pos]^(arr2[pos] = arr2[i]);
                }
            }
        }
        if (!balancedPos.isEmpty())
            System.out.println("No");
        else {
            System.out.println("Yes");
            for (int num : arr2) {
                System.out.print(num + " ");
            }
        }
    }
}