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.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.LinkedList;


public class Party_A115 {
	static ArrayList<Integer>[] adjList;
	static ArrayList<Integer> roots;
  
    static int n;
    public static void main(String[] args) throws NumberFormatException,IOException{
		BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
		 n =Integer.parseInt(br.readLine());
		roots=new ArrayList<>();
		adjList=new ArrayList[n+1];
		for (int i = 0; i < n+1; i++) 
			adjList[i]=new ArrayList<>();
		
		for (int i = 0; i < n; i++) {
			int x=Integer.parseInt(br.readLine());
			if(x!=-1)
		      adjList[x-1].add(i);
			else roots.add(i);
		}
		int max=1;
		for (int i = 0; i < roots.size(); i++) {
			max=Math.max(max, dfs(roots.get(i)));
		}
		System.out.println(max);
	
	}
    static int dfs(int root){
  int  max=1;
  for (int i:adjList[root]) 
	max=Math.max(max, 1+dfs(i));  
	

    
    	
    	return max;
    }
}
