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.*;
import java.util.*;
public class Party {
	static PrintWriter out;
	static boolean vis[];
	static ArrayList<Integer>adj[];
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		out = new PrintWriter(System.out);
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		adj = new ArrayList[n];
		vis = new boolean[n];
		int p;
		for(int i=0;i<n;i++) {
			adj[i] = new ArrayList<>();			
		}
		for(int i=0;i<n;i++) {
			p = sc.nextInt();
			if(p != -1) {
				adj[i].add(p-1);
			}
		}
		Queue<Integer> q = new LinkedList<Integer>();
		for(int i=0;i<n;i++) {
				q.add(i);
		}
		int cnt = 0;
		while(!q.isEmpty()) {
			//out.println(q);
			Queue<Integer> q2 = new LinkedList<>();
			int tmp;
			while(!q.isEmpty()) {
				tmp = q.poll();
				for(int i:adj[tmp]) {
						q2.add(i);
						vis[i] = true;					
				}				
			}
			cnt++;
			q = q2;
		}
		out.println(cnt);
		out.flush();
	}

}
