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<iostream>
#include<algorithm>
#include<map>
#include<cstdio>
#include<cstring>
#include<cmath>
#define ll long long
#define lson (rt<< 1)
#define rson (rt<< 1 | 1)
#define gmid ((l+r)>> 1 )
using namespace std;

const int maxn=20000050;

int a[maxn],b[maxn];

int main()
{
	int n;
	cin>>n;
	for(int i=1;i<=n;++i)
	  cin>>a[i];
	for(int i=1;i<=n;++i)
	  cin>>b[n-i+1];
	int s=0,t=-1;
	for(int i=1;i<=n;++i)
	  if(a[i]==b[i])
	  {
	  	  if(!s) s=i;
	  	  t=i;
	  }     
	int l=1,r=n;  
	for(int i=s;i<=t;++i)
	  {
	  	  if(a[l]^b[i]&&b[l]^b[i]) swap(b[i],b[l++]);
	  	  else if(a[r]^b[i]&&b[r]^b[i]) swap(b[i],b[r--]);
	  	  else {cout<<"No";exit(0);}
	  }  
	cout<<"Yes"<<endl;
	for(int i=1;i<=n;++i)
	  cout<<b[i]<<" ";
	return 0;    
}
