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 <bits/stdc++.h>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
ll n,m,ans;
ll a[200005],b[200005];
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    for(int i=1;i<=n;i++){
        cin>>b[n-i+1];
    }
    ll l=1;
    for(int i=1;i<=n;i++){
		if(a[i]!=b[i])continue;
		while(!(a[i]==b[i]&&a[i]!=a[l]&&b[i]!=b[l])){
			l++;
			if(l>n){
				cout<<"No";
				//system("pause");
				return 0;
			}
		}
		swap(b[l],b[i]);
	}
	cout<<"Yes\n";
	for(int i=1;i<=n;i++)cout<<b[i]<<" ";
    return 0;
}