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 
'''
Created on Sep 10, 2011

@author: dolphinigle
'''

import sys

class Parser:
  def __init__(self):
    self.input = sys.stdin.read().split()
    self.position = 0
  
  def Int(self):
    assert self.position < len(self.input)
    self.position += 1
    return int(self.input[self.position - 1])

cin = Parser()
n = cin.Int()

adj = []
manager = []
fin = []
length = []

for i in range(n):
  adj.append([])
  fin.append(0)
  length.append(0)

for i in range(n):
  manager.append(cin.Int())
  if (manager[-1] != -1):
    adj[manager[-1] - 1].append(i)

daftar = []
idx = 0
for i in range(n):
  if len(adj[i]) == 0:
    daftar.append(i)

ret = 1

while idx < len(daftar):
  node = daftar[idx]
  idx += 1
  length[node] = 1
  for i in adj[node]:
    length[node] = max(length[node], length[i] + 1)
  
  if manager[node] != -1:
    fin[manager[node] - 1] += 1
    if fin[manager[node] - 1] == len(adj[manager[node] - 1]):
      daftar.append(manager[node] - 1)
  ret = max(ret, length[node])

print ret
