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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;

public class Party
{
    static HashMap<Integer, Integer> chain = new HashMap<>();
    static HashMap<Integer, Integer> empMan = new HashMap<>();

    public static void calculate(int emp)
    {
        if(chain.get(emp) != null)
            return;

        int manager = empMan.get(emp);

        if(manager == -1)
        {
            chain.put(emp, 0);
            return;
        }

        if(chain.get(manager) == null)
            calculate(manager);

        int manChain = chain.get(manager);

        chain.put(emp, 1 + manChain);
    }
    public static void main(String [] args)
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        try{
            String line = br.readLine();
            int nEmployees = Integer.parseInt(line);

            for(int i = 0; i < nEmployees; i++)
            {
                line = br.readLine();
                int manager = Integer.parseInt(line);
                empMan.put(i + 1, manager);
            }

            int longest = 0;
            for(int i = 0; i < nEmployees; i++)
            {
                if(chain.get(i + 1) != null)
                    continue;

                calculate(i + 1);
                int length = chain.get(i + 1);

                if(length > longest)
                    longest = length;
            }

            System.out.println(longest + 1);
        }catch(IOException ioe){
            ioe.printStackTrace();
        }
    }
}