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
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.stream.Stream;

public class Main {

  public static void main(String[] args) throws Exception {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    int n = Integer.parseInt(br.readLine());
    int[] a = Stream.of(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
    int[] b = Stream.of(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
    br.close();

    int[] a_cnt = new int[n + 1];
    int[] b_cnt = new int[n + 1];
    int[] c = new int[n + 1];
    int[] d = new int[n + 1];
    for(int i = 0; i < n; i++){
      a_cnt[a[i]]++;
      b_cnt[b[i]]++;
    }

    int max_d = 0;
    String[] array = new String[n];
    for(int i = 1; i <= n; i++){
      if(a_cnt[i] + b_cnt[i] > n){
        System.out.println("No");
        return;
      }
      c[i] = c[i - 1] + a_cnt[i];
      d[i] = d[i - 1] + b_cnt[i];
      max_d = Math.max(max_d, c[i] - d[i - 1]);
    }
    System.out.println("Yes");
    for(int i = 0; i < n; i++){
      array[i] = String.valueOf(b[( i + n - max_d) % n]);
    }
    System.out.println(String.join(" ", array));
  }
}