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.*;

public class party
{
	static Scanner sc=new Scanner(System.in);
	static PrintWriter pw=new PrintWriter(System.out,true);
	static ArrayList<ArrayList<Integer> > gr=new ArrayList<ArrayList<Integer> >();
	static int vis[]=new int[2010];
	static int dfs(int c,int p)
	{
		if (vis[c]==1)
		{
			return 0;
		}
		vis[c]=1;
		int i,z=gr.get(c).size(),ret=1,mx=0;
		for (i=0;i<z;i++)
		{	
			int v=gr.get(c).get(i);
			if (vis[v]==0&&v!=p)
			{
				int got=dfs(v,c);
				if (mx<got)
					mx=got;
			}
		}
		ret=ret+mx; // 1+mx huhuhaaha
		return ret;
	}
	public static void main(String args[])
	{
		int n=sc.nextInt();

		int i,j,p;
		for (i=0;i<=n;i++)
		{
			gr.add(new ArrayList<Integer>());
			vis[i]=0;
		}
		ArrayList<Integer> lolwa=new ArrayList<Integer>();
		for (i=1;i<=n;i++)
		{
			p=sc.nextInt();
			if (p!=-1)
			{
				gr.get(p).add(i);
				gr.get(i).add(p);
			}
			else
			{
				lolwa.add(i);
			}

		}	
		int ans=0;
		for (i=0;i<lolwa.size();i++)
		{
		
			{
				int lol=dfs((int)lolwa.get(i),0);
				if (ans<lol)
					ans=lol;
			}
		}	
		pw.println(ans);

		
	}
}