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.Scanner;


public class Main {

    static int[]d; 
    static int[]g;
    
    public static int getCount(int i)
    {
        if(g[i]==-1)
            return 1;
        else
            return 1+getCount(g[i]);
    }
    
    public static void main(String[] args) 
    {
        Scanner  in =new Scanner(System.in);
        int n = in.nextInt();
        d = new int[n];g = new int[n];
        int i;
        for(i=0;i<n;i++)
        {
            int x = in.nextInt()-1;
            if(x==-2)g[i]=-1;
            else
            {
                g[i] =x;d[x]++;
            }
            
        }
        int max = Integer.MIN_VALUE;
        for(i=0;i<n;i++)
        {
            if(d[i]==0)
                max = Math.max(max, getCount(i));
        }
        System.out.println(max);
    }
}
