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 
numof_employess=int(input())
adjacency_list={}
visited=[False for i in range(numof_employess+1)]
is_root=[False for i in range(numof_employess+1)]
for employee in range(1, numof_employess+1):
    adjacency_list[employee]=[]
for i in range(numof_employess):
    manager=int(input())
    if manager!=-1:
        adjacency_list[manager].append(i+1)
    else:
        is_root[i+1]=True

def get_max_depth_of_tree(start, visited):
    stack=[start]

    max_depth=0
    while stack!=[]:
        V=stack[0]
        max_depth=max(len(stack), max_depth)
        advancing=False
        for neig in adjacency_list.get(V):
            if not visited[neig]:
                stack.insert(0, neig)
                visited[neig]=True
                advancing=True
                break
        if not advancing:
            del stack[0]
    return(max_depth)

global_max_depth=0
for employee in range(1, numof_employess+1):
    if is_root[employee]:
        global_max_depth=max(get_max_depth_of_tree(employee, visited), global_max_depth)
print(global_max_depth)
