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;
#define ll long long
int main()
{
	int n; cin>>n;
	vector<ll>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 i=0,j=n-1;
	bool c=true;
	
	// 1 1 2 3
	// 3 3 2 1
	for(int r=0;r<n;r++)
	{
		if(a[r]==b[r])
		{
			if(a[i]!=b[r] && b[i]!=a[r]) 
			{
				//i++;
				swap(b[i],b[r]);
				i++;
			}
			else if(a[j]!=b[r] && b[j]!=a[r])
			{
				swap(b[j],b[r]);
				j--;
			}
			else
			{
				c=false; break;
			}
			
		}
		
	}
	if(!c)cout<<"No\n";
	else {cout<<"Yes\n";
	for(auto x: b)cout<<x<<" ";}
	
}