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 deque
import sys
def levels(start):
	count=0
	flag=0
	l=[start,-1]
	l=deque(l)
	while l.count(-1)!=len(l):
		element=l.popleft()
		if element==-1:
			count+=1
			l.append(-1)
		else:
			for i in parent[element]:
				l.append(i)
	return count+1
input=sys.stdin.readline
n=int(input())
parent=[[] for i in range(n)]
root=[]
for i in range(n):
	p=int(input())
	if p!=-1:
		parent[p-1].append(i)
	else:
		root.append(i)
m=0
for i in root:
	temp=levels(i)
	if m<temp:
		m=temp
print (m)