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;
typedef long long ll;
const int N = 3e5+10;
const int mod = 1e9+7;
int a[N];
int b[N];
int main() {
	int n;
	cin>>n;
	for(int i=1;i<=n;i++) {
		cin>>a[i];
	}
	for(int i=1;i<=n;i++) {
		cin>>b[i];
	}
	reverse(b+1,b+1+n);
	vector<int>p,q;
	int val=0,cnt=0;
	for(int i=1;i<=n;i++) {
		if(a[i]!=b[i]) continue;
		cnt++;
		p.push_back(i);
		val = b[i]; 
	}
	for(int i=1;i<=n;i++) {
		if(!cnt)break;
		if(a[i]==b[i]) {
			continue;
		}
		if(a[i]!=val&&b[i]!=val) {
			cnt--;
			q.push_back(i);
		}
	}
	if(cnt)  {
	cout<<"No"<<endl;
	return 0;
}
	for(int i=0;i<q.size();i++) {
		swap(b[p[i]],b[q[i]]);
	}
	cout<<"Yes"<<endl;
	for(int i=1;i<=n;i++) {
		cout<<b[i]<<" ";
	}
	
	return 0;
}