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
#include <bits/stdc++.h>
using namespace std;
const int maxn=200005;
int n,L,R,a[maxn],b[maxn];
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	for (int i=1;i<=n;i++) scanf("%d",&b[i]);
	reverse(b+1,b+1+n);
	for (int i=1;i<=n;i++)
		if (a[i]==b[i])
		{
			if (!L) L=i;
			R=i;
		}
	if (L)
	{
		int head=1,x=a[L];
		bool sol=true;
		for (int i=L;i<=R;i++)
		{
			while (head<=n && (a[head]==x || b[head]==x)) head++;
			if (head>n)
			{
				sol=false;
				break;
			}
			swap(b[head++],b[i]);
		}
		if (!sol)
		{
			puts("No");
			return 0;
		}
	}
	puts("Yes");
	for (int i=1;i<n;i++) printf("%d ",b[i]);
	printf("%d\n",b[n]);
	return 0;
}
