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>
#define PI 3.141592653589793238462
using namespace std;
typedef long long ll;
typedef long double db;
ll a[200005],b[200005],p[200005];
int main(){
    ll n;cin>>n;
    for(ll i=1;i<=n;i++){
        cin>>a[i];p[a[i]]++;
    }
    for(ll i=1;i<=n;i++){
        cin>>b[i];p[b[i]]++;
    }
    for(int i=1;i<=n;i++){
        if(p[i]>n){cout<<"No"<<endl;return 0;}
    }
    cout<<"Yes"<<endl;
    reverse(b+1,b+1+n);
    for(int i=1;i<=n;i++){
        if(a[i]==b[i]){
            for(int j=1;j<=n;j++){
                if(b[j]!=a[i]&&b[i]!=a[j]){
                    swap(b[i],b[j]);break;
                }
            }
        }
    }
    for(int i=1;i<=n;i++){
        cout<<b[i]<<" ";
    }cout<<endl;
} 