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 main(){
  int n,i,j,l,r,x;
  cin >> n;
  vector<int> a(n),b(n),ca(n,0),cb(n,0);
  for(i=0;i<n;i++){
    cin >> a.at(i);
    ca.at(a.at(i)-1)++;
  }
  for(i=0;i<n;i++){
    cin >> b.at(i);
    cb.at(b.at(i)-1)++;
  }
  
  for(i=0;i<n;i++){
    if(ca.at(i)+cb.at(i)>n){
      cout << "No" << endl;
      return 0;
    }
  }
  
  i=0;
  j=0;
  
  while(i<n){
    if(a.at(i)==b.at(i)){
      l=i;
      x=a.at(l);
      while(i<n&&a.at(i)==x&&b.at(i)==x){
        i++;
      }
      r=i;
      j=0;
      for(i=l;i<r;i++){
        while(a.at(j)==x||b.at(j)==x){
          j++;
        }
        swap(b.at(i),b.at(j));
      }
      i=r;
    }else{
      i++;
    }
  }
  
  cout << "Yes" << endl;
  for(i=0;i<n;i++){
    cout << b.at(i);
    if(i==n-1){
      cout << endl;
    }else{
      cout << " ";
    }
  }
}