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() {
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  for (auto& e : a) cin >> e;
  for (auto& e : b) cin >> e;

  for (int k = 0; k < 2; k++) {
    for (int i = 0, j = 1; i < n; i++, j = max(j, i + 1)) {
      if (a[i] != b[i]) continue;
      while (j < n && a[i] == b[j]) j++;
      if (j == n) break;
      swap(b[i], b[j]);
    }
    reverse(a.begin(), a.end());
    reverse(b.begin(), b.end());
  }

  // for (auto e : a) cerr << e << ' ';
  // cerr << endl;
  // for (auto e : b) cerr << e << ' ';
  // cerr << endl;

  bool r = true;
  for (int i = 0; i < n; i++) r = r && a[i] != b[i];

  if (r) {
    cout << "Yes" << endl;
    for (auto e : b) cout << e << ' ';
    cout << endl;
  } else {
    cout << "No" << endl;
  }
}
