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 
n = int(input())
p = [int(input()) - 1 for _ in range(n)]


def build_reverse_graph(p):
    graph = {i: [] for i in range(len(p))}
    for i, source in enumerate(p):
        if source != -2:
            graph[source].append(i)
    return graph

def get_level_of_atonement(bitch):
    current, level = bitch, 1
    while p[current] != -2:
        current, level = p[current], level + 1
    return level

def party():
    # print("GRAPH", p)
    reverse_graph = build_reverse_graph(p)
    gods = [i for i in range(len(p)) if p[i] == -2]
    bitches = [i for i in reverse_graph if not reverse_graph[i]]
    # print("REVERSE", reverse_graph)
    # print("GODS", gods)
    # print("BITCHES", bitches)
    # for bitch in bitches:
    #     print(get_level_of_atonement(bitch))
    return max([get_level_of_atonement(bitch) for bitch in bitches])


print(party())
