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<iostream>
#include<cstdio>
#include<queue>
using namespace std;
const int N=1002020;
int n,a[N],b[N];
void print(){
	puts("Yes");
	for(int i=1;i<=n;i++)
		cout<<b[i]<<" ";cout<<endl;
}
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i];
	for(int i=n;i>=1;i--)cin>>b[i];
	int l=n,r=0;
	for(int i=1;i<=n;i++)
		if(a[i]==b[i])l=min(l,i),r=max(r,i);
	if(!r){print();return 0;}
	int nw=l;
	for(int i=1;i<l && nw<=r;i++)
		if(b[nw]!=a[i] && b[nw]!=b[i])
			swap(b[nw],b[i]),++nw;
	for(int i=r+1;i<=n && nw<=r;i++)
		if(b[nw]!=a[i] && b[nw]!=b[i])
			swap(b[nw],b[i]),++nw;
	int fl=0;
	for(int i=1;i<=n;i++)
		if(b[i]==a[i]){fl=1;break;}
	if(fl)puts("No"); else print();
	return 0;
}