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 {


    public static void main(String[] args) {
        Scanner s=new Scanner(System.in);
        int n=s.nextInt();
        
        int emp[]=new int[n];
        for (int i = 0; i < n; i++) {
            emp[i]=s.nextInt()-1;
        }
        
        Queue<Integer>q=new LinkedList<Integer>();
        int max=0;
        
        for (int i = 0; i < n; i++) {
            if(emp[i]==-2){
                q.add(i);
                q.add(1);
            }
        }
        while(!q.isEmpty()){
            int cur=q.poll();
            int dep=q.poll();
            if(dep>max)max=dep;
            for (int i = 0; i < n; i++) {
                if(emp[i]==cur){
                    q.add(i);
                    q.add(dep+1);
                }
            }
            
        }
        System.out.println(max);
    }

}
