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.LinkedList;
import java.util.Queue;
import java.util.Scanner;
public class main22 {
	static class node{
		int data;
		int height;
		node(int d){
			data=d;
			height=1;
		}
	}
	static Queue q = new LinkedList();
	static boolean[] visited;
	static boolean[] group;
	public static void main(String args[]){
		Scanner data = new Scanner(System.in);
		int n = data.nextInt();
		visited = new boolean[n];
		group = visited;
		ArrayList<ArrayList<node>> a = new ArrayList<>(n);
		node[] nd = new node[n];
		for(int i=0;i<n;i++)
			nd[i] = new node(i);
		ArrayList<node> root = new ArrayList<>();
		for(int i=0;i<n;i++)
			a.add(i,new ArrayList<node>());
		for(int i=0;i<n;i++){
			int x = data.nextInt();
			if(x==-1)
				root.add(nd[i]);
			else
				a.get(x-1).add(nd[i]);
		}
		/*for(int i=0;i<n;i++){
			for(int j=0;j<a.get(i).size();j++)
				System.out.print(a.get(i).get(j).data+" ");
				System.out.println();
		}*/
		int max=0;
		for(int i=0;i<root.size();i++){
	//		System.out.println(root.get(i).data);
			int x = bfs(a,root.get(i));
			max=x>max?x:max;
		}
		System.out.println(max);
		
	}
	static int bfs(ArrayList<ArrayList<node>> a,node n){
		q.add(n);
		int max=1;
		while(!q.isEmpty()){
			node x = (node)q.peek();
			for(int i=0;i<a.get(x.data).size();i++){
				if(!visited[a.get(x.data).get(i).data]){
					q.add(a.get(x.data).get(i));
					a.get(x.data).get(i).height=x.height+1;
					if(a.get(x.data).get(i).height>max)
						max=a.get(x.data).get(i).height;
				}
			}
			x = (node)q.remove();
			if(!visited[x.data])
				visited[x.data]=true;
		//	System.out.println(x.data);
		}
		return max;
	}
}
