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 party
{
	static ArrayList[] edges;
	static boolean[] visited;
	static int[] dist;

	static int bfs(int s, int n)
	{
		ArrayList<Integer> q=new ArrayList<Integer>();
		q.add(s);
		visited[s]=true;
		dist[s]=1;

		int levels=1;

		while(!q.isEmpty())
		{
			int u=q.get(0);
			q.remove(0);

			for(int j=0; j<edges[u].size(); j++)
			{
				int v=(int)edges[u].get(j);

				if(!visited[v])
				{
					q.add(v);
					visited[v]=true;
					dist[v]=dist[u]+1;
				}

				if(dist[v]>levels)
					levels=dist[v];
			}
		}

		return levels;
	}

	public static void main(String[] args) {
		Scanner s=new Scanner(System.in);

		int n=s.nextInt();
		edges=new ArrayList[n];
		visited=new boolean[n];
		dist=new int[n];

		ArrayList<Integer> start=new ArrayList<Integer>();

		for(int i=0; i<n; i++)
		{
			edges[i]=new ArrayList<Integer>();
			visited[i]=false;
		}
		for(int i=0; i<n; i++)
		{
			int c=s.nextInt();	
			if(c>0)
				edges[c-1].add(i);
			else
				start.add(i);
		}

		/*for(int i=0; i<n; i++)
		{
			for(int j=0; j<graph[i].size(); j++)
				System.out.print(graph[i].get(j)+" ");

			System.out.println();
		}*/

		int maxlevels=0;
		for(int i=0; i<start.size(); i++)
		{
			int levels=bfs(start.get(i),n);
			if(levels>maxlevels)
				maxlevels=levels;
		}
		System.out.println(maxlevels);
	}
}