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;
const int maxn=4e5+5;
typedef long long ll;
int a[maxn],b[maxn];
int cnt[maxn],n;
int main(){
	cin>>n;
	for (int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		cnt[a[i]]++;
	}
	for (int i=1;i<=n;i++){
		scanf("%d",&b[i]);
		cnt[b[i]]++;
	}
	for (int i=1;i<=n;i++){
		if (cnt[i]>n){
			cout<<"No"<<endl;
			return 0;
		}
	}
	reverse(b+1,b+n+1);
	int l=1,r=n;
	for (int i=1;i<=n;i++){
		if (a[i]==b[i]){
			if (b[i]!=a[l]&&b[l]!=a[i]){
				swap(b[i],b[l]);
				l++;
				continue;
			}
			else if (b[i]!=a[r]&&b[r]!=a[i]){
				swap(b[r],b[i]);
				r--;
				continue;
			}
			cout<<"No"<<endl;
			return 0;
		}
	}
	cout<<"Yes"<<endl;
	for (int i=1;i<=n;i++){
		printf("%d ",b[i]);
	}
}