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.Arrays;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Set;

public class Solution11 {
	static long c = 0;
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		int n = Integer.parseInt(br.readLine());
		
		ArrayList<Integer> al[] = new ArrayList[n + 1];
		ArrayList<Integer> root = new ArrayList<>();
		for(int i=0;i<al.length;i++)
		{
			al[i] = new ArrayList<Integer>();
		}
		
		
		for(int i=0;i<n;i++)
		{
			int a = Integer.parseInt(br.readLine());
			if(a==-1)
			{
				root.add(i+1);
//				al[i+1].add(0);
//				al[0].add(i+1);
			}
			else
			{
				al[i+1].add(a);
				al[a].add(i+1);
			}
		}
//		al[0].add(1);
//		al[1].add(0);
		
		for(int i=0;i<root.size();i++)
		{
			dfs(al,root.get(i),0,1);
		}

		System.out.println(c);
	}
	static void dfs(ArrayList<Integer> al[],int node,int par,int level)
	{
		c = Math.max(c, level);
//		System.out.println("node = "+node+" level ="+level+" c="+c);
		for(int i=0;i<al[node].size();i++)
		{
			if(al[node].get(i)!=par)
			{
				dfs(al,al[node].get(i),node,level+1);
			}
		}
	}

}
