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;
int s=200005,t,l,r,n,a[200005],b[200005];
int main()
{
	scanf("%d",&n);l=1,r=n;
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	for(int i=0;i<n;i++) scanf("%d",&b[n-i]);
	for(int i=1;i<=n;i++)
		if(a[i]==b[i])
		{
			if(s==200005) s=i;
			t=i;
		}
	for(int i=s;i<=t;i++)
		if(a[l]!=b[i]&&b[l]!=b[i]){swap(b[l],b[i]);l++;}
		else if(a[r]!=b[i]&&b[r]!=b[i]){swap(b[r],b[i]);r--;}
		else{puts("No");return 0;}
	puts("Yes");
	for(int i=1;i<=n;i++) printf("%d ",b[i]);
	printf("\n");
	return 0;
}