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 
from collections import defaultdict

class graph:
	def __init__(self,vertex):
		#self.length=vertex
		#self.graph=defaultdict(list)
		self.height=dict()
		self.parent=dict()
		for i in xrange(vertex):
	#		self.graph[i]=[]
			self.height[i]=1
			self.parent[i]=None
	def addedge(self,u,v):
	#	self.graph[u].append(v)
		self.parent[v]=u
		flag=0
		if self.height[u]<(self.height[v]+1):
			self.height[u]=self.height[v]+1
			flag=1
		if flag==1:
			while self.parent[u] is not None:
				self.height[self.parent[u]]=max(self.height[u]+1,self.height[self.parent[u]])
				u=self.parent[u]
	#def __len__(self):
	#	return self.length
"""
def solve1(g,v):
	if len(g.graph[v])==0:
		return 1
	return max((solve1(g,i)+1) for i in g.graph[v])
"""
def solve(g,sup):
	return max(g.height[i] for i in sup)

def main():
	n=input()
	g=graph(n)
	sup=[]
	for i in xrange(n):
		t=input()
		if t==-1:
			sup.append(i)
			continue
		g.addedge(t-1,i)
	count=solve(g,sup)
	print count

if __name__=='__main__':
	main()