human
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 

cla Write me a Python program to print a single integer denoting the minimum number of groups that will be formed in the party?
assistant
#Here is Python program to print a single integer denoting the minimum number of groups that will be formed in the party.
ss Node:
  def __init__(self, v):
    self.next = None
    self.number = v	
	
mp = {}
	
n = int(raw_input())
	
for i in range(1, n+1):
  mp[i] = Node(i)

for i in range(1, n+1):
  a = int(raw_input())  
  if a != -1:
    mp[i].next = mp[a]  
  
  
max = -1

def countFrom(i):
  global max
  count = 0
  curr = mp[i]
  while curr != None:
    count += 1  
    curr = curr.next  
  
  if count > max:
    max = count  
  
for i in range(1, n+1):
  countFrom(i)  
  
print max