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>
#include<cmath>
#include<cstring>
#include<stdio.h>
#include<map>
#include<queue>
#include<vector>
using namespace std;
const int maxn=1e6;

typedef long long ll;

ll a[maxn],b[maxn];

int main(){
	int n;cin>>n;
	for(int i=0;i<n;i++)
		cin>>a[i];
	for(int i=0;i<n;i++)
		cin>>b[i];
	reverse(b,b+n);
	
	//找到相等区间 
	int t=0,l=n,r=-1;
	for(int i=0;i<n;i++){
		if(t==0&&a[i]==b[i]){
			t=a[i];l=r=i;
		}
		else if(t!=0&&a[i]==b[i]){
			r=i;
		}
	}

	//交换后左移 
	for(int i=0;i<n;i++){
		if(a[i]!=t&&b[i]!=t&&l<=r){
			swap(b[i],b[l]);
			l++;
		}
	}
	
	if(l>r){
		cout<<"Yes"<<endl;
        for(int i=0;i<n;i++)
            cout<<b[i]<<" ";
	}
	else cout<<"No"<<endl;
	return 0;
}