A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:

  * Employee A is the immediate manager of employee B
  * Employee B has an immediate manager employee C such that employee A is the superior of employee C. 



The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.

Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.

What is the minimum number of groups that must be formed?

Input

The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.

The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager. 

It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.

Output

Print a single integer denoting the minimum number of groups that will be formed in the party.

Examples

Input

5
-1
1
2
1
-1


Output

3

Note

For the first example, three groups are sufficient, for example: 

  * Employee 1 
  * Employees 2 and 4 
  * Employees 3 and 5 
import java.io.*;

public class Party
{
  static int manager[];
  static int level[];
  public static void main(String[] args) throws IOException
  {
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    int numEmployees = Integer.parseInt(in.readLine());
    manager = new int[numEmployees];
    for(int i = 0; i < numEmployees; i++)
    {
      manager[i] = Integer.parseInt(in.readLine()) - 1;
    }
    int maxGroups = 0;
    level = new int[numEmployees];
    for(int i = 0; i < numEmployees; i++)
    {
      if(manager[i] == -2)
        level[i] = 1;
      else
        findLevel(i);
      if(level[i] > maxGroups)
        maxGroups = level[i];
    }
    System.out.println(maxGroups);
  }
  private static int findLevel(int i)
  {
    if(manager[i] == -2)
    {
      level[i] = 1;
      return 1;
    }
    else if(level[manager[i]] != 0)
    {
      level[i] = level[manager[i]] + 1;
      return level[i];
    }
    else
    {
      level[i] = findLevel(manager[i]) + 1;
      return level[i];
    }
  }
}
