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 a[N],b[N];
int main(){
	int n,si=0,l=N,r,w=0;scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",a+i);
	for(int i=n;i;i--){
		scanf("%d",b+i);
		if(b[i]==a[i])si++,l=i;
	}
	r=l+si-1;
	for(int i=1;i<=n;i++)if(b[i]!=a[r]&&a[i]!=a[r])w++;
	if(w<si)return puts("No"),0;
	puts("Yes");
	for(int i=1;i<=n;i++){
		if(l>r)break;
		if(b[i]!=a[r]&&a[i]!=a[r])swap(b[i],b[l++]);
	}
	for(int i=1;i<=n;i++)printf("%d ",b[i]);
	return 0;
}