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 n,a[400001],b[400001],ta[400001],tb[400001],l=1,r=1;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),ta[a[i]]++;
	for(int i=1;i<=n;i++)scanf("%d",&b[i]),tb[b[i]]++;
	for(int i=1;i<=n;i++)if(ta[i]+tb[i]>n)puts("No"),exit(0);
	memset(b,0,sizeof(b));
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=tb[i];j++)
		{
			while(b[l])++l;
			while(a[r]<=a[l]&&r<=n)++r;
			while(b[r])++r;
			if(a[l]!=i)b[l++]=i;
			else b[r++]=i;
		}
	}
	l=1;
	for(int i=n+1;b[i];i++)
	{
		while(b[l])++l;
		if(a[l]!=b[i])a[l]=b[i];
		else 
		{
			while(a[r=(rand()%n+1)]==b[i]||b[r]==a[l]);
			b[l]=b[r],b[r]=b[i];
		}
	}
	puts("Yes");
	for(int i=1;i<=n;i++)printf("%d ",b[i]);
	return 0;
}
