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{
    static int max=0;
    public static void main(String args[]) {
        Scanner scn=new Scanner(System.in);
        int n=Integer.parseInt(scn.nextLine());
        ArrayList<Integer>[]adj=new ArrayList[n+1];
        for(int i=0;i<=n;i++){
            adj[i]=new ArrayList<>();
        }
        for(int i=1;i<=n;i++){
            int parent=Integer.parseInt(scn.nextLine());
            if(parent==-1){adj[0].add(i);}
            else{adj[parent].add(i);}
        }
        dfs(0,adj,0);
        System.out.println(max);
    }
    public static void dfs(int src,ArrayList<Integer>adj[],int level){
        max=Math.max(level,max);
        for(int nei:adj[src]){
            dfs(nei,adj,level+1);
        }
    }
}