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 <algorithm>
#include <cstdint>
#include <iostream>
#include <vector>

using namespace std;

int main() {
  int64_t N;
  cin >> N;
  vector<int64_t> As(N+1);
  vector<int64_t> Bs(N);
  vector<int64_t> Ds(N+1);
  for (int64_t n = 0; n < N + N; ++n) {
    int64_t i;
    cin >> i;
    if (n < N) {
      ++As[i];
    } else {
      Bs[n % N] = i;
      ++Ds[i];
    }
  }
  for (int64_t n = 1; n <= N; ++n) {
    if (As[n] + Ds[n] > N) {
      cout << "No" << endl; return 0;
    }
    As[n] += As[n-1];
    Ds[n] += Ds[n-1];
  }
  cout << "Yes" << endl;
  
  int64_t mx = 0;
  for (int64_t n = 1; n <= N; ++n) {
    mx = max(mx, As[n] - Ds[n-1]);
  }
  for (int64_t n = 0; n < N; ++n) {
    if (n) cout << " ";
    cout << Bs[(n + N - mx) % N];
  }
  cout << endl;

  return 0;
}
