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.InputStreamReader;
import java.util.LinkedList;
import java.io.*;

@SuppressWarnings("unused")
public class problemC {

	static LinkedList<Integer> [] children;
	public static int solve(int idx, int height){
		if(children[idx].size() == 0)return height;
		int max = 0;
		for(int i = 0 ; i < children[idx].size() ; ++i){
			max = Math.max(max, solve(children[idx].get(i), height + 1));
		}
		return max;
	}
	public static void main(String[] args)throws IOException {
		
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int N = Integer.parseInt(br.readLine());
		children = new LinkedList [N + 1];
		for(int i = 0 ; i < children.length ; ++i)
			children[i] = new LinkedList<Integer>();
		for(int i = 0 ; i < N ; ++i){
			int parent = Integer.parseInt(br.readLine());
			if(parent == -1){
				children[0].add(i + 1);
			}else{
				children[parent].add(i + 1);
			}
		}
		int ans = 0;
		for(int i = 0 ; i < children[0].size() ; ++i){
			int st = children[0].get(i);
			ans = Math.max(ans, solve(st, 1));
		}
		System.out.println(ans);
		
	}
}
