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 
class Node:
	def __init__(self, index):
		self.manager = None
		self.sup = set()
		self.index = index

	def add_manager(self, manager):
		self.manager = manager

	def get_superiors(self):
		#print(self.manager, self.sup)
		if len(self.sup) == 0:
			if self.manager is None:
				return set()
			self.sup = self.manager.get_superiors()
			self.sup.add(self.manager)
		return self.sup.copy()
		
	def do_get_superiors(self): 
		if len(self.sup) == 0 and self.manager is not None: self.get_superiors()

	def in_superior(self, others):
		return len(others & self.sup) > 0

	def is_superior(self, other):
		return other in self.sup
	def __repr__(self):
		return str(self)+": manager: "+str(self.manager)+", sup: {"+", ".join([str(s) for s in self.sup])+"}"

	def __str__(self): return str(self.index)+"M"
	def __eq__(self, obj):
		return obj.index == self.index
	def __hash__(self): return self.index
ls = [Node(_) for _ in range(int(input()))]
for i in range(len(ls)):
	x = int(input())
	if x != -1: ls[i].add_manager(ls[x-1])
m = 0
for l in ls: m = max(m, len(l.get_superiors()))
print(m+1)
