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.ArrayList;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Party {
	
	static ArrayList<Integer>[] graph;
	
	static int counter = 1;
	static int max = 1;
	
	@SuppressWarnings("unchecked")
	public static void main(String[] args) {
		// TODO Auto-generated method stub

		
		Scanner in = new Scanner(System.in);
		
		int n = in.nextInt();
		
		graph = new ArrayList[n];
		
		Set<Integer> superiors = new HashSet<Integer>();
		
		for(int i = 0; i< n ; ++i) {
			int a = in.nextInt();

			if(a != -1) {
				if(graph[a-1] == null) graph[a-1] = new ArrayList<Integer>();
				graph[a-1].add(i);
				if(graph[i] == null) graph[i] = new ArrayList<Integer>();
			}
			else {
				if(graph[i] == null) graph[i] = new ArrayList<Integer>();
				superiors.add(i);
			}
								
		}
		in.close();


		for(Integer i: superiors) {
			max = Math.max(max, DFS(i));			
		}
		
		System.out.println(max);
		
	}
	
	public static int DFS(int n) {
		
		int ans = 1;
		int temp = 0;
		for(int i = 0; i < graph[n].size(); ++i ) {
			temp = Math.max(temp,DFS(graph[n].get(i)));
		}
		return ans+temp;
	}
	
}
