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.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;

/**
 * Created by simin on 3/5/15.
 */
public class C
{
    public static final int max = 2000 + 7;
    public static void main(String []strings) throws IOException
    {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter writer = new PrintWriter(new OutputStreamWriter(System.out));
        int n = Integer.parseInt(reader.readLine());
        ArrayList<Integer> []adjacencyList = new ArrayList[max];
        for (int i = 0; i < max; i++)
            adjacencyList[i] = new ArrayList<Integer>();
        Queue<Integer> queue = new LinkedList<Integer>();
        int []best = new int[max];
        Arrays.fill(best, -1);
        for (int i = 0; i < n; i++)
        {
            int v = Integer.parseInt(reader.readLine());
            if (v != -1)
                adjacencyList[v - 1].add(i);
            else
            {
                best[i] = 1;
                queue.offer(i);
            }
        }
        while (!queue.isEmpty())
        {
            int top = queue.poll();
            for (int i = 0; i < adjacencyList[top].size(); i++)
            {
                queue.offer(adjacencyList[top].get(i));
                best[adjacencyList[top].get(i)] = best[top] + 1;
            }
        }
        int res = 0;
        for (int i = 0; i < n; i++)
            res  = Math.max(res, best[i]);
        writer.println(res);
        writer.flush();
        writer.close();
    }
}
