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;
typedef pair<int,int> P;
int main(){
    int n;
    cin >> n;
    vector<int> a(n),b(n);
    for(int i=0;i<n;i++)cin>>a[i];
    for(int i=0;i<n;i++)cin>>b[i];
    vector<int> s(n+1),t(n+1);
    for(int i=0;i<n;i++){
    	s[a[i]]++;
		t[b[i]]++;
	}
    for(int i=0;i<=n;i++)
    if(s[i]+t[i]>n){
        cout<<"No"<<endl;
        return 0;
    }
    for(int i=0;i<n;i++){
    	s[i+1]+=s[i];
		t[i+1]+=t[i];
	}
    int x=-2e5-10;
    for(int i=0;i<n;i++)x=max(x,s[i+1]-t[i]);
    cout<<"Yes"<<endl;
    for(int i=0;i<n;i++){
    	cout<<b[(i+n-x)%n];
		if(i==n-1)cout<<endl;
		else cout<<' ';
	}
    return 0;
}