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=2e5+10;
inline int qr()
{
	int x=0,f=0;
	char ch=0;
	while(!isdigit(ch))
	{
		f=ch=='-';
		ch=getchar();
	}
	while(isdigit(ch))
	{
		x=(x<<1)+(x<<3)+(ch^48);
		ch=getchar();
	}
	return f?-x:x;
}
int n;
int a[maxn],b[maxn];
int p[maxn];
int cnt;
int val;
int main()
{
	n=qr();
	for(int i=1;i<=n;i++)
	{
		a[i]=qr();
	}
	for(int i=1;i<=n;i++)
	{
		b[i]=qr();
	}
	reverse(b+1,b+n+1);
	for(int i=1;i<=n;i++)
	{
		if(b[i]!=a[i])continue;
		cnt++;
		p[cnt]=i;//相等的位置;
		val=b[i]; 
	}
	for(int i=1;i<=n;i++)
	{
		if(!cnt) break;
		if(a[i]!=val&&b[i]!=val)//可以交换 
		swap(b[p[cnt--]],b[i]);
	}
	if(cnt) //位置不够 
	{
	printf("No");return 0;	
	}
	else {
		printf("Yes\n");
		for(int i=1;i<=n;i++)
		{
			printf("%d ",b[i]);
		}
	}
	return 0;
}