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

num_emps = int(input())
emp_managers = [int(input()) for x in range(num_emps)]

emp_tree = []
root_nodes = []


for i in range(len(emp_managers)):
    emp_tree.append({
        'value': i + 1,
        'children': []
    })

for i in range(len(emp_managers)):
    if emp_managers[i] != -1:
        emp_tree[emp_managers[i] - 1]['children'].append(i)
    else:
        root_nodes.append(i)

# def depth(node):
#     if len(emp_tree[node]['children']) == 0:
#         return 1
#     else:
#         return 1 + max([depth(node) for node in emp_tree[node]['children']])


def depth(root_node):
    depth = 0
    q = deque([root_node])
    child_count = 1
    next_row_count = 0
    while q:
        node = q.popleft()

        child_count -= 1
        
#         print('visiting '  + str(node))
#         print('child count '  + str(child_count))
        
        q += emp_tree[node]['children']
        next_row_count += len(emp_tree[node]['children'])
        
        if child_count == 0:
            depth += 1
            child_count = next_row_count
            next_row_count = 0
            
    return depth
        
    
max_depth = max([depth(node) for node in root_nodes])   
    
# print(emp_tree)
# print(root_nodes)
print(max_depth)
