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, value):

		self.value = value
		self.depth = 0
n=int(input())
p=[]
d={}

for i in range(n):
	k=int(input())
	if k!=-1:
		if k not in d:
			d[k] = [i+1]
		else:
			d[k].append(i+1)
	else:
		p.append(i+1)
visited = [False]*n
l = [node(i) for i in range(1,n+1)]
dep = 0
ans=0
for s in p:
	if s in d:
		queue = [] 
		queue.append(s)

		while queue:
			s=queue.pop(0)
			dep = l[s-1].depth
			if visited[s-1]==False:
				visited[s-1] = True
				l[s-1].depth = dep+1
				dep=l[s-1].depth

			ans=max(ans,dep)

			if s in d:
				for i in d[s]:
					if visited[i-1] == False:
						queue.append(i)
						visited[i-1] = True
						l[i-1].depth = dep+1
				dep = l[s-1].depth+1
					
			ans=max(ans,dep)
if ans==0:
	ans+=1	
print(ans)