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 
''' codeforces '''
from Queue import *
class node(object):
	''' atmost one immediate manager'''
	def __init__(self,label):
		self.label =label
		self.manager=None
		self.junior=[]
		self.colour='white'
		
n=int(raw_input())
store=[]
for i in range(n):
	store.append(node(i))
for i in range(n):
	s=int(raw_input())
	if s!=-1:
		store[i].manager=store[s-1]
		store[s-1].junior.append(store[i])

# NEXT STEP: FIND THE TREE WITH THE HIGHEST HEIGHT 
def height(node):
	if node.colour!='white':
		return
	while node.manager!=None:
		node=node.manager
	q=Queue()
	q.put(node)
	level=0
	while q.empty() != True:
		level+=1
		out=q.qsize()
		while out>0:
			current=q.get()
			for i in current.junior:
				if i.colour=='white': 
					i.colour='gray'
					q.put(i)
			out-=1
	return level
h=height(store[0])
for node in store:
	h_new=height(node)
	if h_new>h:
		h=h_new
print h

	







 