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;

public class CF_115 {
	
	static ArrayList<Integer>[]adjList;
	static int currDepth;
	
	public static void main(String[] args) throws NumberFormatException, IOException {
		
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		ArrayList<Integer>roots=new ArrayList<>();
		int n=Integer.parseInt(br.readLine());
		adjList=new ArrayList[n];
		for(int i=0;i<n;i++)
			adjList[i]=new ArrayList<>();
		for(int i=0;i<n;i++)
		{
			int parent=Integer.parseInt(br.readLine());
			if(parent==-1)
				roots.add(i);
			else
				adjList[parent-1].add(i);
		}
		int maxDepth=0;
		for(int i=0;i<roots.size();i++)
		{
			currDepth=0;
			int currRoot=roots.get(i);
			dfs(currRoot,1);
			if(currDepth>maxDepth)
				maxDepth=currDepth;
		}
		System.out.println(maxDepth);
	}
	public static void dfs(int u,int depth)
	{
		if(adjList[u].size()==0)
		{
			if(depth>currDepth)
				currDepth=depth;
			return;
		}
		for(int v:adjList[u])
			dfs(v,depth+1);

	}
	
	

}
