human
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 

clas
assistant
#Here is Python code to print a single integer denoting the minimum number of groups that will be formed in the party.
s hierarchy():
    lookup = dict()
    def __init__(self,num,manager_num,manager=None):
        self.manager = manager
        self.manager_num = manager_num
        self.num = num
    def get_level(self):
        if self.manager == None:
            return 0
        else:
            count = 0
            superior = self.manager
            while superior != None:
                count += 1
                superior = superior.manager
            return count
employee_num = int(input())
employees = []
for x in range(employee_num):
    manager_num = int(input())
    emp_obj = hierarchy(x,manager_num)
    employees.append(emp_obj)
    hierarchy.lookup[x+1] = emp_obj
for emp in employees:
    emp.manager = hierarchy.lookup.get(emp.manager_num)
levels = set([emp.get_level() for emp in employees])
print(len(levels))