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 
# https://codeforces.com/problemset/problem/115/A

from collections import defaultdict, deque

def construct_graphs(n, array):

    graph_dict = defaultdict(list)
    level_dict = dict()
    sources = []

    for idx, superior_num in enumerate(array):
        if superior_num > 0:
            graph_dict[superior_num - 1].append(idx)

        else:
            sources.append(idx)

    return sources, graph_dict


def DFS(sources, graph_dict):

    heights = []
    for source in sources:

        level_dict = dict()
        level_dict[source] = 1
        max_height = 1
        queue = deque()
        queue.append(source)

        while len(queue) != 0:

            node = queue.popleft()
            for employee in graph_dict[node]:

                max_height = level_dict[node] + 1
                level_dict[employee] = max_height
                queue.append(employee)

        heights.append(max_height)

    return max(heights)


if __name__ == '__main__':

    n = int(input())
    array = []
    for __ in range(n):
        array.append(int(input()))

    sources, graph_dict = construct_graphs(n, array)
    print(DFS(sources, graph_dict))






