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 c116C
{
    private final Scanner sc;
    private static final boolean debug = true;

    static void debug(Object ... objects)
    {
        if(debug)
            System.err.println(Arrays.toString(objects));        
    }

    c116C()
    {
        sc = new Scanner(System.in);
    }

    public static void main(String [] args)
    {
        (new c116C()).solve();
    }

    int [] parent, depth;

    
    void solve()
    {
        int n = sc.nextInt();
        parent= new int[n];
        depth = new int[n];
        Arrays.fill(depth,-1);
        HashMap<Integer, ArrayList<Integer> > g = new HashMap<Integer, ArrayList<Integer>>();
        LinkedList<Integer> Q = new LinkedList<Integer>();

        for(int i=0;i<n;i++)
        {
            if(!g.containsKey(i))
                g.put(i, new ArrayList<Integer>());

            int j = sc.nextInt();
            if(j>0)j--;
            parent[i] = j;
            if(parent[i] == -1)
            {
                depth[i] = 0;
                Q.add(i);
            }
            else
            {
                if(!g.containsKey(parent[i]))
                    g.put(parent[i],new ArrayList<Integer>());
                g.get(parent[i]).add(i);
            }                
        }

        int maxDepth = 0;
        while(Q.size() > 0)
        {
            int a  = Q.getFirst();            
            Q.removeFirst();
            for(int i:g.get(a))
            {
                depth[i] = depth[a] + 1;
                if(maxDepth < depth[i])
                    maxDepth = depth[i];
                Q.add(i);
            }
        }

        System.out.println(maxDepth + 1);
        

    }
}
