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.*;

public class CodeForces{
    public static void main(String args[])throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine().trim());
        Node arr[] = new Node[n+1];
        for(int i = 0 ; i <= n ; i++){
            arr[i] = new Node(i);
        }
        int emp = 1;
        while(n-- > 0){
            int boss = Integer.parseInt(br.readLine().trim());
            if(boss == -1)
                boss = 0;
            arr[boss].addEmployee(arr[emp++]);
        }
        arr[0].print(arr[0] , 0);
        System.out.println(arr[0].maxDepth);
    }
}
class Node{
    int value;
    ArrayList<Node> emps;
    int maxDepth = 0;
    Node(int value){
        this.value = value;
        emps = new ArrayList<Node>();
    }
    public void addEmployee(Node e){
        this.emps.add(e);
    }
    public ArrayList<Node> getEmployees(Node e){
        return e.emps;
    }
    public void print(Node n , int depth){
        if(depth > this.maxDepth){
            this.maxDepth = depth;
        }
        for(Node i : n.emps){
            print(i , depth +1);
        }
    }
}