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+5;
int n,k,a[N],f[N],b[N],F,le,ri,l,r;
int main(){
	cin>>n;
	for(k=1;k<=n;k++){
		cin>>a[k];
		f[a[k]]++;
	}
	for(k=1;k<=n;k++){
		cin>>b[k];
		f[b[k]]++;
		if(f[b[k]]>n){
		F=1;
		} 
	}
	if(F) cout<<"No"<<endl;
	else
	{
		reverse(b+1,b+n+1);
		for(k=1;k<=n;k++){
			if(a[k]==b[k])F=a[k];
		}
	    
		if(F){ l=n;
			for(k=1;k<=n;k++){
				if(a[k]==F) l=min(l,k);
				if(b[k]==F) l=min(l,k);
			}le=1; ri=n;
			for(k=1;k<=n;k++){
				if(b[k]==F && a[k]==F){
					if(le<l){
					swap(b[k],b[le]);le++;
					}
					else {
						swap(b[k],b[ri]); 
						ri--;
					}
				}
			}
		}
		cout<<"Yes"<<endl;
		for(k=1;k<=n;k++){
			cout<<b[k]<<" ";
		}
	}
}