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.ArrayList;
import java.util.Scanner;


public class A {

    /**
     * @param args
     */
    static int[] a;
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner sc = new Scanner(System.in);
        int n= sc.nextInt();
        a = new int[n];
        int[] b = new int[n];
        int k=0;
        Node[] nodes = new Node[n];
        for (int i=0; i<n; i++)
            nodes[i] = new Node(i);
        for (int i=0; i<n; i++){
            int p = sc.nextInt()-1;
            if (p==-2) {
                b[k++] = i;
            }
            else
                nodes[p].add(i);
            
        }
        int counter = 0;
        for (int i=0; i<k; i++){
            int p = dfs(nodes, b[i]);
            if (p>counter) counter=p;
        }
        System.out.println(counter);
        
    }
    public static int dfs(Node[] nodes, int i){
        if (nodes[i].size()==0) return 1;
        int k = 0;
        for (int j=0; j<nodes[i].size(); j++){
            int p = dfs(nodes,nodes[i].get(j));
            if (p>k) k =p; 
        }
        return k+1;
    }
    

}
class Node{
    public int index;
    ArrayList<Integer> lst = new ArrayList<Integer>();
    public int get(int i){
        return lst.get(i);
    }
    public int size(){
        return lst.size();
    }
    public void add(int p){
        lst.add(p);
    }
    public Node(int index){
        this.index = index;
    }
}