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.BufferedReader;

import java.io.PrintWriter;

import java.io.InputStreamReader;
import java.util.*;
public class Main{
 static int n,t;
static int f[];
 
 public static void  main(String args[]){ 
     Scanner a = new Scanner(new BufferedReader(new InputStreamReader(System.in)));

     while(a.hasNext()){
      int n=a.nextInt();
      f=new int[n];
      for(int i=0;i<n;i++)f[i]=a.nextInt()-1;
      int ans=0;
      for(int i=0;i<n;i++){
    	  int h=0,w=i;
    	  while(f[w]!=-2){w=f[w];h++;}
    	  ans=Math.max(h, ans);
      }
      System.out.println(ans+1);
     }
 }  
}   