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 Main
{
    private static Scanner scan = new Scanner(System.in);
    
    public static void main(String[] args)
    {
        int n=scan.nextInt();
        graph G = new graph(n);
        int i=1;
        while(i<=n)
        {
            int j=scan.nextInt();
            if(j!=-1)
            {
                //System.out.println("From" + j + " To" + i);
                G.addedge(j,i);
            }
            i++;
        }
        int ans=0;
        for(i=1;i<=n;i++)
        {
            int j=G.dfs(i);
            j++;
            if(j>ans)
                ans=j;
        }
        System.out.println(ans);
    }
}

class graph
{
    public ArrayList<ArrayList<Integer>> grph ;
    graph(int val)
    {
    grph = new ArrayList<ArrayList<Integer>>(val+1);//list of nodes an list of array list
        for(int i=0;i<=val;i++)
        {
            ArrayList<Integer> arr =new ArrayList<Integer>();
            grph.add(arr);
        }
    }
    public void addedge(int from, int to)//function to add an edge integer is passed
    {
        (grph.get(from)).add(to);
    }
    
    public int dfs(int i)
    {
        //System.out.println(i);
        Iterator<Integer> itr = (grph.get(i)).iterator();
        int max=0;
        while(itr.hasNext())
        {
            int temp=dfs(itr.next())+1;
            if(temp>max)
                max=temp;
        }
        return max;
    }
}
