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<algorithm>
using namespace std;
int N_MAX = 200001;
int main()
{
  int n,i,j,c=0;
  int a[N_MAX],b[N_MAX],ca[N_MAX],cb[N_MAX];
  cin >> n;
  fill(ca+1,ca+n+1,0);
  fill(cb+1,cb+n+1,0);
  for(i=0;i<n;i++){
    cin >> a[i];
    ca[a[i]]++;
  }
  for(i=0;i<n;i++){
    cin >> b[i];
    cb[b[i]]++;
  }
  for(i=1;i<=n;i++)
    if(ca[i]+cb[i]>n){
      cout << "No" << endl;
      return 0;
    }
  cout << "Yes" << endl;
  for(i=0;i<n;i++){
    if(a[i]!=b[i])
      continue;
    if(b[i]!=c){
      j=0;
      c=b[i];
    }
    for(;j<n;j++)
      if(a[j]!=c&&b[j]!=c){
        b[i]=b[j];
        b[j]=c;
        j++;
        break;
      }
  }
  for(i=0;i<n;i++)
    cout << b[i] << " ";
  cout << endl;
  return 0;
}