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;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N;
  cin >> N;
  vector<int> A(N), B(N), C(N + 10, 0), D(N + 10, 0);
  for (auto&& e : A) {
    cin >> e;
    C[e]++;
  }
  for (auto&& e : B) {
    cin >> e;
    D[e]++;
  }
  for (int i = 0; i < N + 10; i++) {
    if (C[i] + D[i] > N) {
      cout << "No" << '\n';
      return 0;
    }
  }
  for (int i = 0; i < N; i++) {
    C[i + 1] += C[i], D[i + 1] += D[i];
  }
  int d = 0;
  for (int i = 0; i < N; i++) {
    d = max(d, D[i + 1] - C[i]);
  }
  cout << "Yes" << '\n';
  for (int i = 0; i < N; i++) {
    cout << B[(i + d) % N] << (i == N - 1 ? '\n' : ' ');
  }
  return 0;
}
