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.*;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;

public class Party {
    static ArrayList<Integer>[] list;
    static boolean[] visited;
    static int[] distance;
    static ArrayList<Integer> source;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        list = new ArrayList[n];
        visited = new boolean[n];
        distance = new int[n];
        source = new ArrayList<>();
        for(int i = 0; i < n; i++){
            list[i] = new ArrayList<>();
        }
        for(int i = 0; i < n; i++){
            st = new StringTokenizer(br.readLine());
            int A = Integer.parseInt(st.nextToken()) - 1;
            if(A != -2) {
                list[i].add(A);
                list[A].add(i);
            } else{
                source.add(i);
            }
        }
        int maxDepth = 0;
        for(int i : source){
            bfs(i);
        }
        for(int i = 0; i < n; i++){
            if(distance[i] > maxDepth)
                maxDepth = distance[i];
        }
        pw.println(maxDepth);
        pw.close();
    }

    public static void bfs(int sourceNode){
        Queue<Integer> q = new LinkedList<>();
        q.add(sourceNode);
        visited[sourceNode] = true;
        distance[sourceNode] = 1;
        while(!q.isEmpty()){
            int n = q.remove();
            visited[n] = true;
            for(int i = 0; i < list[n].size(); i++){
                if(!visited[list[n].get(i)]){
                    q.add(list[n].get(i));
                    distance[list[n].get(i)] = distance[n] + 1;
                }
            }
        }
    }
}