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=200010;
int a[N],b[N],n;
 
int main(){
	scanf("%d",&n);
	int x;
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	for(int i=1;i<=n;i++) scanf("%d",&b[i]);
	reverse(b+1,b+1+n);
	int l=0,r=0,t=0,L=0,R=0;
	for(int i=1;i<=n;i++) if(a[i]==b[i]) {
		if(!l) l=i,r=i,t=a[i];
		else r++;
	}
	if(!l){
		printf("Yes\n");
		for(int i=1;i<=n;i++) printf("%d ",b[i]);
		return 0;
	}
	for(int i=1;i<=n;i++) if(a[i]==t || b[i]==t) {L=i-1;break;}
	for(int i=n;i>=1;i--) if(a[i]==t || b[i]==t) {R=i+1;break;}
	if(L+n-R+1<(r-l+1)) printf("No\n");
	else{
		for(int i=1;i<=min(L,r-l+1);i++) swap(b[i],b[l+i-1]);
		for(int i=1;i<=r-l+1-min(r-l+1,L);i++) swap(b[n-i+1],b[r-i+1]);
		printf("Yes\n");
		for(int i=1;i<=n;i++) printf("%d ",b[i]);
	}
}