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.util.*;

public final class Party
{static int l=0;
	static boolean marked[];
	static ArrayList<Integer>[] arr;
	static ArrayList<Integer> list;
	public static void main(String[] args) {
		Scanner br=new Scanner(System.in);
		int n=br.nextInt();
		marked=new boolean[n+1];
		arr=new ArrayList[n+1];
		list=new ArrayList<Integer>();
		ArrayList<Integer> free=new ArrayList<Integer>();
		int ans=0;
		for(int i=0;i<=n;i++)
			arr[i]=new ArrayList<Integer>();
		for(int i=1;i<=n;i++)
		{
			int s=br.nextInt();
			if(s==-1)
				free.add(i); //person who has no manager
			else
			arr[s].add(i); //s is manager of i
		}
		for(Integer i:free)
			if(arr[i].size()>0)
				{
					bfs(i); //person who is superior of someone but doesn't have supirior is the root
				ans=Math.max(l,ans);
				l=0;
				}
		System.out.println(ans+1);

	}
	public  static void bfs(int s)
{
	int temp=1,c=0;
	ArrayDeque<Integer> q=new ArrayDeque<Integer>();
	marked[s]=true;
	q.offerLast(s);
	while(!q.isEmpty())   
	{
		int v=q.pollFirst();
		temp--;
	for(int w:arr[v]) 
	{
		if(!marked[w])			
			{
				
				marked[w]=true;
				q.offerLast(w);
				c++;
				list.add(w);
			}
	}
	if(temp==0)
	{
	temp=c;
	c=0;
	if(list.size()>0)
	l++;
	
	
	list.clear();
	
	}
}

}

}