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.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;

public class Party {

    private static List<Set<Integer>> groups = new ArrayList<Set<Integer>>();
    private static Map<Integer, Set<Integer>> s = new HashMap<Integer, Set<Integer>>();

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();

        int[] emp = new int[n];
        int i = 0;

        while (i < n) {
            emp[i++] = scanner.nextInt();
        }
        groups.add(new HashSet<Integer>());
        
        fillMap(emp);
        
        for(i = 0 ; i < n ; i++) {
            boolean inserted = false;
            for(Set<Integer> group : groups) {
                boolean found = false;
                for(Integer em : group) {
                    if(s.get(i).contains(em) || s.get(em).contains(i)) {
                        found = true;
                        break;
                    }
                }
                if(!found){
                    group.add(i);
                    inserted = true;
                    break;
                }
                
            }
            if(!inserted) {
                Set<Integer> s1 = new HashSet<Integer>();
                s1.add(i);
                groups.add(s1);
            }
        }
        
        System.out.println(groups.size());
    }

    private static void fillMap(int[] emp) {
        for (int i = 0; i < emp.length; i++) {
            if (!s.keySet().contains(i)) {
                s.put(i, new HashSet<Integer>());
            }
            int j = emp[i];
            while (j != -1) {
                j -=1;
                s.get(i).add(j);
                j = emp[j];
            }
        }

    }

}
