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 
'''
Created on 2/12/2014

@author: Samuel Cabezas
'''
def make_link(G, node1, node2):
    if node1 not in G:
        G[node1] = {}
    (G[node1])[node2] = 1
    if node2 not in G:
        G[node2] = {}
    return G

def DFS_iterative(G, node):
    color = {}
    distance_from_start = {}
    for v in G:
        color[v] = 'white'
    color[node] = 'gray'
    nodelist = [node]
    distance_from_start[node] = 1
    while nodelist <> []:
        u = nodelist.pop()
        if u in G:
            for neighbor in G[u].keys()[::-1]:
                if neighbor <> 0:
                    if color[neighbor] == 'white':
                        color[neighbor] = 'gray'
                        nodelist.append(neighbor)
                        distance_from_start[neighbor] = distance_from_start[u] + 1
            color[u] = 'black'
    return distance_from_start

n = int(raw_input())
line = []
G = {}
roots = []
maximo = []
for i in range(n):
    line.append(int(raw_input()))
    if line[i] <> -1:
        make_link(G, line[i], i+1)
    else:
        make_link(G,i+1,0) 
        roots.append(i+1)
for x in roots:
    a = DFS_iterative(G, x)
    maximo.append(max(a.values()))

print max(maximo)

