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+5;
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[i];
    
    reverse(b+1,b+n+1);
    
    //for(int i=1;i<=n;++i)
	//    cout<<b[i]<<" ";
		
	//cout<<'\n'; 
    
    int flag=-1;
    int l=1000,r=-1;

    for(int i=1;i<=n;++i) 
	{
        if(flag==-1&&a[i]==b[i]) 
		{ 
            flag=a[i];
            l=i;r=i;
        } 
        
		else if(flag!=-1&&a[i]==b[i]) r=i;
    }
    
    for(int i=1;i<=n;++i) 
	{
        if(flag!=a[i]&&flag!=b[i]&&l<=r){
            swap(b[i],b[l]);
            l++;
        }
    }
 
    if(l<=r){
        cout<<"No"<<'\n'; 
    } 
    
	else{
        cout<<"Yes"<<'\n';
        for(int i=1;i<=n;++i) {
            cout<<b[i]<<" ";
        }
    }
 
    return 0;
}