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 <iostream>
#include <string>
#include <math.h>
#include <algorithm>
#include <vector>
using namespace std;
int main(){
  int N;
  cin>>N;
  int A[200000],B[200000];
  for(int i=0;i<N;i++) cin>>A[i];
  for(int i=0;i<N;i++) cin>>B[i];
  sort(B,B+N,greater<int>());
  vector<int> same;
  for(int i=0;i<N;i++){
    if(A[i]==B[i]) same.push_back(i);
  }
  int j=0;
  int judge=0;
  for(int i=0;i<same.size();i++){
    while(A[same[i]]==B[same[i]]){
      if(j==N){
        judge++; break;
      }
      if(A[j]!=B[same[i]]&&B[j]!=B[same[i]]) swap(B[same[i]],B[j]);
      j++;
    }
    if(judge==1) break;
  }
  if(judge==0){
    cout<<"Yes"<<endl;
    for(int i=0;i<N-1;i++) cout<<B[i]<<' ';
    cout<<B[N-1]<<endl;
  }
  else cout<<"No"<<endl;
  
  return 0;
}