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<algorithm>

using namespace std;

const int N = 2*1e5;

int n,x[N],y[N];

bool cmp(int a,int b){
	return a>b;
}

int main(){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d",&x[i]);
	}
	for(int i=0;i<n;i++){
		scanf("%d",&y[i]);
	}
	sort(y,y+n,cmp);
	int flag=0;
	for(int i=0;i<n;i++){
		if(x[i]==y[i]){
			flag=i;
			break;
		}
	}
	while(x[flag]==y[flag]){
		for(int i=0;i<n;i++){
			if(x[i]!=x[flag] && y[i]!=x[flag]){
				swap(y[i],y[flag]);
				flag++;
				break;
			}
			if(i==n-1){
				printf("No");
				return 0;
			}
		}
	}
	printf("Yes\n");
	for(int i=0;i<n;i++){
		printf("%d ",y[i]);
	}
	return 0;
}