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 
# The task boils down to finding the largest number of levels in a connected component of a DAG.
import collections


def num_levels(graph, s, visited):
    visited[s] = True

    q = [s]

    res = 1
    while True:
        q_tmp = []
        for v in q:
            for u in graph[v]:
                if not visited[u]:
                    q_tmp.append(u)

        if not q_tmp:
            break
        q = q_tmp
        res += 1

    return res


def max_num_levels(graph, roots, n):
    visited = [False] * n

    max_levels = 1
    for root in roots:
        num_levels_root = num_levels(graph, root, visited)
        if num_levels_root > max_levels:
            max_levels = num_levels_root
    return max_levels


n = int(input())
roots = []
graph = [[] for _ in range(n)]
for child in range(n):
    parent = int(input())
    if parent == -1:
        roots.append(child)
    else:
        graph[parent - 1].append(child)

print(max_num_levels(graph, roots, n))
