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

@SuppressWarnings("unchecked")
public class Main {
    private static final long MOD = (int)1e9 + 7;

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] a = new int[n];
        int[] b = new int[n];

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

        int x = -1;
        int maxI = Integer.MIN_VALUE;
        int minI = Integer.MAX_VALUE;
        for (int i = 0; i < n; ++i) {
            if (a[i] == b[i]) {
                minI = Math.min(minI, i);
                maxI = Math.max(maxI, i);
                x = a[i];
            }
        }

        int i = 0;
        int idx = minI;
        if (idx != Integer.MAX_VALUE) {
           while (i < n) {
            if (idx >= minI && idx <= maxI && a[i] != x && b[i] != x) {
                int temp = b[i];
                b[i] = x;
                b[idx++] = temp;
            } 
            ++i;
        } 
        }
        

        for (i = 0; i < n; ++i) {
            if (a[i] == b[i]) {
                System.out.println("No");
                return;
            }
        }
        System.out.println("Yes");
        for (i = 0; i < n; ++i) {
            System.out.print(b[i] + " ");
        }
    }
}