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<cstring>
#include<algorithm>
using namespace std;
#define N 200020
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,a[N],b[N],d,cnt[N];
int main(){
	n=read();
	for(int i=1;i<=n;++i){
		a[i]=read();
		++cnt[a[i]];
	}
	for(int i=1;i<=n;++i){
		b[i]=read();
		++cnt[b[i]];
	}
	int pa=1,pb=1;
	for(int i=1;i<=n;++i){
		while(a[pa]==i)++pa;
		d=max(d,pa-pb);
		while(b[pb]==i)++pb;
	}
	rotate(b+1,b+n+1-d,b+n+1);
	for(int i=1;i<=n;++i){
		if(a[i]==b[i])return !printf("No\n");
	}
	printf("Yes\n");
	for(int i=1;i<=n;++i){
		printf("%d ",b[i]);
	}
	return 0;
}

