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<stdio.h>
void swap(int&x,int&y){int t=x;x=y;y=t;}
const int Maxn=200010;
int n,a[Maxn],b[Maxn],c[Maxn],d[Maxn],ld=0;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),c[a[i]]++;
	for(int i=1;i<=n;i++)scanf("%d",&b[i]),c[b[i]]++;
	for(int i=1;i<=n;i++)if(c[i]>n)return puts("No"),0;
	puts("Yes");
	for(int i=1;i<=n/2;i++)swap(b[i],b[n-i+1]);
	int L=-1,R,x;
	for(int i=1;i<=n;i++)
	if(a[i]==b[i])
	{
		L=i;x=a[i];
		for(int j=i;j<=n;j++)
		if(a[j]==b[j])R=j;
		break;
	}
	if(L!=-1)
	{
		for(int i=1;i<=n;i++)
		if(a[i]!=x&&b[i]!=x)d[++ld]=i;
		for(int i=L;i<=R;i++)
		swap(b[i],b[d[ld--]]);
	}
	for(int i=1;i<=n;i++)printf("%d ",b[i]);
 } 