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 <iostream>
#include <vector>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;

int main(void) {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<int> A(N);
  vector<int> Acnt(N + 1, 0);
  rep(i, N) {
    cin >> A[i];
    Acnt[A[i]]++;
  }
  vector<int> B(N);
  vector<int> Bcnt(N + 1, 0);
  rep(i, N) {
    cin >> B[i];
    Bcnt[B[i]]++;
  }
  rep(i, N + 1) {
    if (N < (Acnt[i] + Bcnt[i])) {
      cout << "No" << endl;
      return 0;
    }
  }

  int j = 0;
  int prev = -1;
  rep(i, N) {
    if (prev != A[i]) j = 0;
    if (A[i] == B[i]) {
      for (; j < N; ++j) {
        if (A[i] != A[j] && A[i] != B[j]) {
          swap(B[i], B[j]);
          break;
        }
      }
    }
    prev = A[i];
  }

  cout << "Yes" << endl;
  rep(i, N) {
    if (i != 0) cout << " ";
    cout << B[i];
  }
  cout << endl;
  return 0;
}
