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;
template <typename T> bool chmax(T &u, const T z) { if (u < z) {u = z; return true;} else return false; }
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

int main() {
  int n;
  cin>>n;
  vector<int> a(n), b(n);
  rep(i,n)cin>>a[i];
  rep(i,n)cin>>b[i];
  sort(b.rbegin(),b.rend());
  map<int,int> c;
  rep(i,n)c[a[i]]++;
  rep(i,n)c[b[i]]++;
  int saidai=0;
  for(auto p:c)chmax(saidai,p.second);
  if (saidai > n) {
      cout<<"No"<<endl;
      return 0;
  }
  cout<<"Yes"<<endl;
  rep(i,n){
    if(a[i]!=b[i])continue;
    else{
        rep(j,n){
          if(a[i]!=b[j]&&a[j]!=b[i]){
            swap(b[i],b[j]);
            break;
          }
        }
    }
  }
  rep(i,n)cout<<b[i]<<" ";
  cout<<endl;
  return 0;
}