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 
from collections import deque

class Node(object):
    def __init__(self, ind, parent_ind):
        self.id = ind
        self.children = []
        self.parent_ind = parent_ind
        self.parent = None
        self.height = -1

    def add_parent(self, node):
        self.parent = node
        node.children.append(self)

    def set_height(self, height = 1):
        self.height = height
        
        q = deque()
        q.append(self)

        while len(q) > 0:
            node = q.popleft()
            for child in node.children:
                child.height = node.height + 1
                q.append(child)

n = int(raw_input())
nodes = [None]*(n + 1)
for i in range(1, n + 1):
    nodes[i] = Node(i, int(raw_input()))

for i in range(1, len(nodes)):
    if nodes[i].parent_ind != -1:
        nodes[i].add_parent(nodes[nodes[i].parent_ind])

try:
    for i in range(1, len(nodes)):
        if nodes[i].parent is None:
            nodes[i].set_height(1)
    
    max_depth = max(node.height for node in nodes if node is not None)

    print max_depth 
except Exception as e:
    print e