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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.StringTokenizer;

public class Party {


    static boolean [] visited;
    static ArrayList<Integer> [] adj;



    public static int bfs(int v){

        LinkedList<Integer>queue = new LinkedList<>();
        visited[v] = true;
        queue.add(v);
        int count = 1;

        LinkedList<Integer> next = new LinkedList<>();

        while(queue.size()!=0) {



            int size = queue.size();
            for(int j = 0 ; j < size ; j++) {
                int point = queue.pollFirst();
                for (int i = 0; i < adj[point].size(); i++) {
                    if (!visited[adj[point].get(i)]) {
                        next.add(adj[point].get(i));
                        visited[adj[point].get(i)] = true;
                    }
                }
            }

            if(next.size()!=0)count++;
            queue = next;
        }




        return count;

    }




    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int n = Integer.parseInt(br.readLine())+1;

        visited = new boolean[n];
        adj = new ArrayList[n];

        for(int i = 0 ; i<n ; i++){
            adj[i] = new ArrayList<>();
        }


        for(int i = 1 ; i<n ; i++){

            StringTokenizer st = new StringTokenizer(br.readLine());
            while (st.hasMoreTokens()){
                int manager = Integer.parseInt(st.nextToken());
                if(manager!=-1)
                    adj[manager].add(i);
            }
        }



        int max = Integer.MIN_VALUE;

        for(int i = 1 ; i<n ; i++){

            int tmp = bfs(i);
            if (tmp>max){
                max = tmp;
            }

            visited = new boolean[n];

        }

        System.out.println(max);



    }




}
