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)
using namespace std;
using ll = long long;
using P = pair<int,int>;


int main() {
  int n;
  cin >> n;

  vector<int> A(n), B(n), C(n + 1), D(n + 1);
  rep(i, n) {
    int a;
    cin >> a;
    A[i] = a;
    C[a]++;
  }
  rep(i, n) {
    int b;
    cin >> b;
    B[i] = b;
    D[b]++;
  }

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

  rep(i, n) C[i + 1] += C[i];
  rep(i, n) D[i + 1] += D[i];

  int x = 0;
  rep(i, n) x = max(x, C[i + 1] - D[i]);

  cout << "Yes" << endl;
  rep(i, n) {
     cout << B[(i + n - x) % n] << ' ';
  }
  cout << endl;
  
  return 0;
}
