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 N=2e5+10;
int a[N],b[N];
int n;
int t[N],t1[N];
inline int read()
{
	int Num=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9') {if(ch=='-') f=-1; ch=getchar();}
	while(ch>='0'&&ch<='9') {Num=(Num<<1)+(Num<<3)+ch-'0'; ch=getchar();}
	return Num*f;
}
int tot=200,flag=0;
int main()
{
	n=read();int Bit=1,_bit=n;
	for(int i=1;i<=n;i++) a[i]=read();
	for(int i=1;i<=n;i++) b[n-i+1]=read();
	for(int i=1;i<=n;i++) 
	if(a[i]==b[i])  
	{
		if(b[Bit]!=a[i]&&b[i]!=a[Bit]) 
			swap(b[Bit++],b[i]);
		else if(b[_bit]!=a[i]&&a[_bit]!=b[i])
			swap(b[_bit--],b[i]);
		else 
		{
			puts("No");
			return 0;
		}
	}
		puts("Yes");
		for(int i=1;i<=n;i++)
		printf("%d ",b[i]);
		return 0;
	return 0;
}