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 
# def fun(gh,cur):
# 	print 'fun:',cur
# 	global ARR
# 	ARR[cur]=1
# 	if len(gh[cur]) == 0:
# 		print 'return 0'
# 		return 0
# 	m=0
# 	for i in gh[cur]:
# 		t=fun(gh,i)
# 		if t>m:
# 			m=t
# 	return m+1

roots=[]
def fun(gh,cur):
	floor=1
	queue=[(cur,floor)]
	#print queue
	while queue:
		x=queue.pop(0)
		if len(gh[x[0]])>0:
			floor = x[1]+1
		for t in gh[x[0]]:
			queue.append((t,floor))
		#print queue
		#print floor
	return floor
	
n=input()
gh=[[] for _ in xrange(n+1)]
for i in xrange(1,n+1):
	p=input()
	if p != -1:
		gh[p].append(i)
	else:
		roots.append(i)
max=0
for i in roots:
	t=fun(gh,i)
	if t>max:
		max=t
print max
