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>

#define gc() getchar()

#define ri register int 

using std::max;

inline int read()
{
	int ret=0;char ch=gc();
	while(!isdigit(ch))ch=gc();
	while(isdigit(ch))
	{
		ret=ret*10+ch-'0';
		ch=gc();
	}
	return ret;
}

const int MAXN=2e5+10;

int n,d;
int a[MAXN],b[MAXN],c[MAXN];
int l[MAXN],r[MAXN];

int main()
{
//	freopen("in.in","r",stdin);
	
	n=read();
	
	memset(r,-1,sizeof(r));
	memset(l,-1,sizeof(l));
	for(ri i=0;i<n;++i)
	{
		a[i]=read();
		r[a[i]]=i;
	}
	for(ri i=0;i<n;++i)
	{
		b[i]=read();
		if(l[b[i]]==-1)l[b[i]]=i;
	}
	
	for(ri i=1;i<=n;++i)if(l[i]!=-1 && r[i]!=-1)
		d=max(d,r[i]-l[i]+1);
	
	for(ri i=0;i<n;++i)
	{
		c[(i+d)%n]=b[i];
		if(a[(i+d)%n]==c[(i+d)%n])
		{
			printf("No\n");
			return 0;
		}
	}
	
	printf("Yes\n");
	for(ri i=0;i<n;++i)printf("%d ",c[i]);
	
	return 0;
}