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
const int N=2e5+5;
int n,m;
int a[N], b[N], c[N];
int main(){
	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);
	int l, r, flag, num;
	l = r = flag = num = 0;
	for(int i = 1; i <= n; i ++ ){
	 if(a[i]==b[i]&&l==0)
	    {
	    	l=i,r=i,flag=a[i];
		}
		else  
		   if(a[i]==b[i]&&l)
		   {
		   	  r=i;
		   }
	}
	if(l){
		for(int i = 1; i <= n && num <= r - l; i ++ ){
			if(b[i] != flag && a[i] != flag){
				b[l + num] = b[i];
				b[i] = flag;
				num ++ ;
			}
		} 
	}
	if(l == 0 || num == r - l + 1){
		puts("Yes");
		for(int i = 1;i <= n; i ++ ){
			cout << b[i] << " ";
		}
		cout << endl;
	}
	else{
		puts("No");
	}
	return 0;
}