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
assistant
#Here is Python program to print a single integer denoting the minimum number of groups that will be formed in the party.

from sys import stdin

class Node:
    def __init__(self):
        self.val = 1
        self.parent = None

n = int(stdin.readline())
a = [Node() for _ in range(n)]
res = 1
for i in range(n):
    t = int(stdin.readline())
    if t != -1:
        a[i].parent = a[t - 1]
        a[i].val = a[t - 1].val + 1
        res = max(res, a[i].val)
changed = True
while changed:
    changed = False
    for i in range(n):
        if a[i].parent is not None and a[i].val != a[i].parent.val + 1:
            a[i].val = a[i].parent.val + 1
            res = max(res, a[i].val)
            changed = True
print(res)