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 license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
//package javaapplication11;

import java.util.Scanner;

/**
 *
 * @author Josepe
 */
public class JavaApplication11 {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int min=0;
        int[] un=new int[n+1];
        for (int i = 1; i < n+1; i++) {
            int l=in.nextInt();
            if(l==-1)
                un[i]=i;
            else
                un[i]=l;
            
        }
        
        for (int i = 1; i <=n; i++) {
            if(find(un,i)>min)
                min=find(un,i);
        }
        System.out.println(min);
        
    }
    
    //----------------------------------------------
    public static int find(int[] un,int i){
        if(un[i]==i) return 1;
        else return 1+find(un,un[i]);
    }
}
