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 
//package CodeForce;

import java.util.*;

import java.io.*;
public class round115A {

	
	public static void main(String[] ard)  throws IOException
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw=new PrintWriter(System.out);
		int n=Integer.valueOf(br.readLine().trim());
		LinkedList<Integer> l[]=new LinkedList[n];
		for(int i=0;i<n;i++)
			l[i]=new LinkedList<Integer>();
		Queue<Integer> q=new LinkedList<Integer>();
	     int[]	depth=new int[n];
		for(int i=0;i<n;i++)
		{
			int a=Integer.valueOf(br.readLine().trim());
			if(a!=-1)
			{
				l[a-1].add(i);
			}
			else 
			{
				depth[i]=1;
				q.offer(i);
			}
		}
		while(!q.isEmpty())
		{
			int u=q.poll();
			for(int g:l[u])
			{
				q.offer(g);
				depth[g]=depth[u]+1;
			}
		}
		int max=0;
		for(int i=0;i<n;i++)
			max=Math.max(max, depth[i]);
		pw.println(max);
			
			
	pw.close();
		
	}
	 
}

