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 < (int)(n); i++)
using namespace std;

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

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

  reverse(b.begin(),b.end());
  rep(i,n + 1) {
    if(cnt[i] > n) {
      cout << "No" << '\n';
      return 0;
    }
  }
  int x = 0;
  int k = 0;
  vector<int> overlap;
  for (int i = 0; i < n; i++) {
    if(a[i] == b[i]) {
      x = a[i];
      overlap.push_back(i);
      k++;
    }
  }
  vector<int> idx;
  for (int i = 0; i < n; i++) {
    if(a[i] != x && b[i] != x) {
      idx.push_back(i);
    }
  }

  for (int i = 0; i < k; i++) {
    swap(b[overlap[i]], b[idx[i]]);
  }

  cout << "Yes" << '\n';
  for (auto it : b) cout << it << " ";

}
