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.*;
public class Main {
    public static void DFS(int[]arr,boolean[]visited,int[]dist,int i)
    {
        visited[i]=true;
        if (arr[i]==-1)
        {
            dist[i]=1;
            return;
        }
        if (!visited[arr[i]])
            DFS(arr,visited,dist,arr[i]);
        dist[i]=dist[arr[i]]+1;
        return;
    }
    public static void main(String[] args) {
        int n;
        Scanner sc=new Scanner(System.in);
        n=sc.nextInt();
        int[]arr=new int[n+1];
        for (int i=1;i<n+1;i++)
            arr[i]=sc.nextInt();
        boolean[]visited=new boolean[n+1];
        int[]dist=new int[n+1];
        for (int i=1;i<n+1;i++)
            visited[i]=false;
        for (int i=1;i<n+1;i++)
        {
            if (arr[i]==-1) {
                dist[i] = 1;
                visited[i] = true;
            }
            else
            {
                if (!visited[i])
                  DFS(arr,visited,dist,i);
            }
        }
        int res=Arrays.stream(dist).max().getAsInt();
        System.out.println(res);
    }
}