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 
// package Ladder14;

import java.util.*;

public class CF115A {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        int[] arr = new int[n];
        HashMap<Integer,ArrayList<Integer>> adjacent = new HashMap<>();
        HashSet<Integer> vertices = new HashSet<>();
        for (int i = 0; i < n; i++) {
            vertices.add(i+1);
        }
        for (int i = 0; i < n; i++) {
            int k = s.nextInt();
            if(k != -1) {
                if(adjacent.containsKey(k)) {
                    ArrayList<Integer> l1 = adjacent.get(k);
                    l1.add(i+1);
                    adjacent.put(k,l1);
                }else{
                    ArrayList<Integer> l1 = new ArrayList<>();
                    l1.add(i+1);
                    adjacent.put(k,l1);
                }
                vertices.remove(i+1);
            }
        }

//        System.out.println(adjacent.toString());
//        System.out.println(vertices.toString());

        Iterator<Integer> list = vertices.iterator();
        long max = Integer.MIN_VALUE;
        while(list.hasNext()){
            int curr = list.next();
            int ans = calculateHeight(curr, adjacent,0);
            if(max < ans + 1){
                max = ans + 1;
            }
        }
        System.out.println(max);
    }

    private static int calculateHeight(int curr, HashMap<Integer, ArrayList<Integer>> adjacent, int level) {
        ArrayList<Integer> list = adjacent.get(curr);
        if(list == null){
            return level;
        }
        int max = level;
        for (int i = 0; i < list.size(); i++) {
            int k = calculateHeight(list.get(i),adjacent,level  +1 );
            if(k > max){
                max = k;
            }
        }
        return max;

    }
}