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.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

public class RENAMETHISBITCH {
	
	public static void main(String[] args) {
		
		try (Scanner sc = new Scanner(System.in)) {
			
			int n = sc.nextInt();
			List<Integer> employees = new ArrayList<>();
			Map<Integer, Node> map = new HashMap<>();
			List<Integer> layer = new ArrayList<>();
			
			map.put(-1, new Node());
			layer.add(-1);
			
			for (int i = 0; i < n; i++) {
				employees.add(sc.nextInt());
			}
			
			while (map.size() < employees.size() + 1) {
				
				for (int i = 0; i < n; i++) {
					if (map.containsKey(employees.get(i))) {
						Node x = new Node();
						map.get(employees.get(i)).add(x);
						map.put(i + 1, x);
					}
				}
			}
			
			System.out.println(map.get(-1).getLength(0));
		}
		catch (Exception e) {
			e.printStackTrace();
		}
	}
}

class Node {
	List<Node> nexts;
	
	public Node() {
		nexts = new ArrayList<>();
	}
	
	public void add(Node n) {
		nexts.add(n);
	}
	
	public int getLength(int l) {
		
		if (nexts.isEmpty()) return l;
		else {
			int max = l + 1;
			for (Node next : nexts) {
				max = Math.max(max, next.getLength(l + 1));
			}
			return max;
		}
	}
}