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 class Solution
{
	
    static ArrayList<Integer>[] graph;
    static int [] levelSingle ;
    static int [] levels;
    static boolean [] isVis;
    
    
    static void bfs(int source)
    {
        Queue<Integer> myQueue =new LinkedList<>();
        int level =1;
         myQueue.add(source);
        isVis[source] =true;
        levelSingle[source] =1;
        levels[1]++;
        while(!myQueue.isEmpty())
        {
            int temp = myQueue.remove();
            
            for(Integer i:graph[temp])
            {
                
                    myQueue.add(i);
                    isVis[i] = true;
                    levelSingle[i] =levelSingle[temp]+1;
                    levels[levelSingle[i]]++;
                
                
            }

            
        }
    
    }
    public static void main(String [] args)
    {
        
        Scanner in = new Scanner(System.in);
        
  
        
       
            int n = in.nextInt();
            
            graph =new ArrayList[n+1];
            for(int j =0;j<=n;j++)
            {
                graph[j]  =new ArrayList<>();
            }
            
            for(int j =1;j<=n;j++)
            {
                int parent =in.nextInt();
                
                if(parent!=-1)
                {
                    graph[parent].add(j);
                }
            }
            
             levels =new int [n+1];
             levelSingle =new int [n+1];
             isVis =new boolean[n+1];
             
             for(int j=1;j<=n;j++)
             {
            	 
                 bfs(j);
             }
            
            int ans=0;
            for(Integer single :levels)
            {
            	
                if(single>0)
                ans++;
            }
            System.out.println(ans);
            
        }
        
        
    
    
    
    
}
