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 queue import Queue


def minimun_groups(graph, ceos):
    depths = [0] * len(graph)
    queue = Queue()
    for ceo in ceos:
        queue.put(ceo)
        while not queue.empty():
            boss = queue.get()
            for employee in graph[boss]:
                depths[employee] = depths[boss] + 1
                queue.put(employee)
    print(len(set(depths)))


def main():
    total_employees = int(input())
    employees = []
    ceos = []
    for i in range(total_employees):
        employees.append(int(input()))
        if employees[i] == -1:
            ceos.append(i)

    graph = [[] for _ in employees]
    for i, boss in enumerate(employees):
        if boss != -1:
            graph[boss - 1] += [i]
    minimun_groups(graph, ceos)


if __name__ == "__main__":
    main()
