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 maxn=200010;
int n,c=-1,l=0,r=-1,a[maxn]={},b[maxn]={},h[maxn]={};
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)scanf("%d",&b[i]),h[n-i+1]=b[i];
	for(int i=1;i<=n;i++)b[i]=h[i];
	for(int i=1;i<=n;i++)if(a[i]==b[i]){c=a[i];break;}
	for(int i=1;i<=n;i++)if(a[i]==c&&b[i]==c){l=i;break;}
	for(int i=n;i>=1;i--)if(a[i]==c&&b[i]==c){r=i;break;}
	for(int i=1;i<=n;i++)if(a[i]!=c&&b[i]!=c&&l<=r)swap(b[l],b[i]),l++;
	if(l<=r)printf("No");
	else{printf("Yes\n");for(int i=1;i<=n;i++)printf("%d ",b[i]);}
	return 0;
}