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;

int main() {
  int N,Z=0;
  cin>>N;
  vector<int> p(N);
  vector<int> q(N);
  vector<int> r(N);
  for(int i=0;i<N;i++){
    cin>>p[i];
    r[p[i]-1]++;
    Z=max(Z,r[p[i]-1]);
  }
  for(int i=0;i<N;i++){
    cin>>q[i];
    r[q[i]-1]++;
    Z=max(Z,r[q[i]-1]);
  }
  if(Z>N){
    cout<<"No"<<endl;
    return 0;
  }
  cout<<"Yes"<<endl;
  reverse(q.begin(),q.end());
  int A=0,B=N-1;
  for(int i=0;i<N;i++){
    if(p[i]!=q[i]){
      continue;
    }
    if(p[i]!=q[A]&&p[A]!=q[i]){
      swap(q[A],q[i]);
      A++;
    }
    else{
      swap(q[B],q[i]);
      B--;
    }
  }
  for(int i=0;i<N;i++){
    cout<<q[i];
    if(i+1==N){
      cout<<endl;
    }
    else{
      cout<<" ";
    }
  }
}