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 
import sys

class Emp:
  parent = -1
  depth = 0
  children = []

n = int(sys.stdin.readline().strip())

emps = []
for i in xrange(n+1):
  e = Emp()
  e.children = e.children[:]
  emps.append(e)
  
for i in xrange(1,n+1):
  p = int(sys.stdin.readline().strip())
  if p < 0:
    p = 0
  emps[i].parent = p
  emps[p].children.append(i)
 
for i in xrange(0,n+1):
  if len(emps[i].children) == 0:
    emps[i].depth = 1
    current = i
    while (current != -1) and (emps[current].parent != -1):
      next = emps[current].parent
      if emps[next].depth < emps[current].depth + 1:
        emps[next].depth = emps[current].depth + 1
        current = next
      else:
        current = -1

res = emps[0].depth - 1
print(res)