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;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1001001001;
const int MOD = 1e9 + 7;

int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> b(n);
  rep(i, n) cin >> a[i];
  rep(i, n) cin >> b[i];
  reverse(b.begin(), b.end());
  int lhr = 0;
  int rhr = b.size() - 1;
  rep(i, b.size()) {
    if (a[i] == b[i]) {
      if (a[lhr] != b[i] && b[lhr] != b[i]) {
        int t = b[lhr];
        b[lhr] = b[i];
        b[i] = t;
        lhr++;
      } else if (a[rhr] != b[i] && b[rhr] != b[i]) {
        int t = b[rhr];
        b[rhr] = b[i];
        b[i] = t;
        rhr--;
      } else {
        cout << "No" << endl;
        return 0;
      }
    }
  }
  cout << "Yes" << endl;
  rep(i, n - 1) cout << b[i] << " ";
  cout << b[n - 1] << endl;
  return 0;
}