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 
/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */


import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;

/**
 *
 * @author sousnake
 */
public class C116_Party {
    public static void main(String as[]) throws IOException{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        int p[]=new int[n+1];
        for(int i=1;i<=n;i++){
            p[i] = Integer.parseInt(br.readLine());
        }
        Queue<Integer> q= new LinkedList<Integer>();
        boolean vis[]=new boolean[n+1];
        int level[] =new int[n+1];
        for(int i=1;i<=n;i++){
            if(p[i]==-1)
                q.add(i);
        }
        //for(int i=0;i<=n;i++)
        //    System.out.println("he "+p[i]);
        //System.out.println(q);
        while(!q.isEmpty()){
            int x = q.poll();
            vis[x]=true;
            for(int i=1;i<=n;i++){
                if(p[i]==x&&vis[i]==false){
                    //System.out.println("here "+i+" "+p[i]);
                    q.add(i);
                }
            }
            //System.out.println(x+" "+p[x]);
            if(p[x]==-1)
                level[x]=1;
            else level[x]=level[p[x]]+1;
        }
        int max=1;
        for(int i=1;i<=n;i++){
            if(max<level[i])
                max=level[i];
        }
        System.out.println(max);
    }
}
