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;
template<class C>constexpr int sz(const C&c){return int(c.size());}
using ll=long long;constexpr const char nl='\n',sp=' ';
 
int main() {
  // freopen("in.txt", "r", stdin);
  // freopen("out.txt", "w", stdout);
  // freopen("err.txt", "w", stderr);
  ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  int N;
  cin >> N;
  vector<int> A(N), B(N);
  for (auto &&a : A) cin >> a;
  for (auto &&b : B) cin >> b;
  int mx = INT_MIN;
  for (int i = 0; i < N; i++) mx = max(mx, int(upper_bound(B.begin(), B.end(), A[i]) - B.begin() - i));
  if (mx < 0) mx += N;
  rotate(B.begin(), B.begin() + mx, B.end());
  for (int i = 0; i < N; i++) if (A[i] == B[i]) {
    cout << "No" << nl;
    return 0;
  }
  cout << "Yes" << nl;
  for (int i = 0; i < N; i++) cout << B[i] << " \n"[i == N - 1];
  return 0;
}
