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 ll long long int
#define mod 1000000007
using namespace std;

int main(){
	
	int n;
	cin>>n;
	vector<int>a(n),b(n);
	for(int i=0;i<n;i++){
		cin>>a[i];
	}
	for(int i=0;i<n;i++){
		cin>>b[i];
	}
	reverse(b.begin(),b.end());
	int lpt=0,rpt=n-1;
	for(int i=0;i<n;i++){
		
		if(a[i]==b[i]){
			
			if(b[lpt]!=b[i] && a[lpt]!=b[i]){
				swap(b[i],b[lpt]);
				++lpt;
			}
			else if(b[rpt]!=b[i] && a[rpt]!=b[i]){
				swap(b[i],b[rpt]);
				--rpt;
			}
			else{
				cout<<"No\n";
				return 0;
			}
		}
	}
	for(int i=0;i<n;i++){
		
		if(a[i]==b[i]){
			cout<<"No\n";
			return 0;
		}
	}
	cout<<"Yes\n";
	for(auto x:b){
		cout<<x<<" ";
	}
		cout<<endl;
	return 0;
}
