Problem: 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

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ios ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int MAXN=2e5+100;
int a[MAXN],b[MAXN];
int main(){
	ios;
	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);
	int l=INF,r=-1,c;
	for(int i=1;i<=n;i++){
		if(a[i]==b[i]){
			c=a[i];
			l=min(l,i);
			r=max(r,i);
		}
	}
	for(int i=1;i<=n;i++){
		if(l>r) break;
		if(l<=r && b[i]!=c && a[i]!=c){
			swap(b[i],b[l]);
			l++;
		}
	}
	if(l>r){
		cout<<"Yes"<<'\n';
		for(int i=1;i<=n;i++){
			if(i!=n) cout<<b[i]<<' ';
			else cout<<b[i]<<'\n';
		}
	}
	else cout<<"No"<<'\n';
	return 0;
}