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>
#define rep(i,n) for (int i = 0; i < (n); i++)
#define sz(x) int(x.size())
using namespace std;
typedef long long ll;
const int inf = 1001001001;

int main(){
  int n;
  cin >> n;
  vector<int> a(n), b(n), aa(n+1,0), bb(n+1,0);
  rep(i,n) {
    cin >> a[i];
    aa[a[i]]++;
  }
  rep(i,n) {
    cin >> b[i];
    bb[b[i]]++;
  }

  rep(i,n){
    if(bb[i+1]+aa[i+1] > n){
      cout << "No" << endl;
      return 0;
    }
  }


  cout << "Yes" << endl;
  rep(i,n){
    aa[i+1] += aa[i];
    bb[i+1] += bb[i];
  }
  
  int x = 0;
  rep(i,n){
    x = max(x,aa[i+1]-bb[i]);
  }
  rep(i,n){
    cout << b[(i-x+n)%n] << " ";
  }
  cout << endl;


}
