Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main(){
  int N;cin>>N;
  vector<int>A(N+1);
  vector<int>B(N+1);
  vector<int>C(N);
  vector<int>D(N);
  for(int i=0;i<N;i++){
    cin>>C[i];
    A[C[i]]++;
  }
  for(int i=0;i<N;i++){
    cin>>D[i];
    B[D[i]]++;
  }
  for(int i=1;i<N+1;i++){
    if(A[i]+B[i]>N){
      puts("No");return 0;
    }
  }
  puts("Yes");
  for(int i=0;i<N;i++){
    A[i+1]+=A[i];
    B[i+1]+=B[i];
  }
  int count=0;
  for(int i=0;i<N;i++)count=max(count,A[i+1]-B[i]);
  for(int i=0;i<N;i++)cout<<D[(i+N-count)%N]<<(i!=N? " ":"");
  cout<<"\n";
}