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 n;
vector<int> a, b, res;

bool solve();

int main() {
  cin >> n;
  a.resize(n);
  for (auto &p : a) cin >> p;
  b.resize(n);
  for (auto &p : b) cin >> p;
  if (solve()) {
    cout << "Yes" << endl;
    for (int i = 0; i < n; ++i) cout << res[i] << " \n"[i == n - 1];
  } else
    cout << "No" << endl;
  return 0;
}

bool solve() {
  vector<int> cnta(n + 1, 0), cntb(n + 1, 0);
  for (int i = 0; i < n; ++i) {
    ++cnta[a[i]];
    ++cntb[b[i]];
  }
  for (int i = 1; i <= n; ++i) {
    if (cnta[i] + cntb[i] > n) return 0;
    cnta[i] += cnta[i - 1];
    cntb[i] += cntb[i - 1];
  }
  int dif = 0;
  for (int i = 1; i <= n; ++i) dif = max(dif, cnta[i] - cntb[i - 1]);
  res.resize(n);
  for (int i = 0; i < n; ++i) res[(i + dif) % n] = b[i];
  return 1;
}