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;
#define   ll long long int
#define maxn 200005








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];
reverse(b.begin(),b.end());
int el=-1;
for(int i=0;i<n;i++){
    if(a[i]==b[i]){
        el=a[i];
    }
}
if(el==-1){
    cout<<"Yes"<<endl;
    for(auto x: b)
        cout<<x<<" ";
    cout<<endl;
}
else{
queue<int> q1;
queue<int> q2;
for(int i=0;i<n;i++){
    if(a[i]==el && b[i]==el)
        q2.push(i);
    if(a[i]!=el && b[i]!=el){
        q1.push(i);
    }
}
if(q2.size()>q1.size()){
    cout<<"No"<<endl;
}
else{
cout<<"Yes"<<endl;
while(!q2.empty()){
    int i1=q2.front();
    q2.pop();
    int i2=q1.front();
    q1.pop();
    swap(b[i1],b[i2]);
}
for(auto x: b)
    cout<<x<<" ";

cout<<endl;
}
}

}
