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 
class CodeforcesTask115ASolution:
    def __init__(self):
        self.result = ''
        self.n = 0
        self.superior = []

    def read_input(self):
        self.n = int(input())
        for x in range(self.n):
            self.superior.append(int(input()))

    def process_task(self):
        graph = [[[], 1] for x in range(self.n)]
        for x in range(self.n):
            if self.superior[x] > 0:
                graph[self.superior[x] - 1][0].append(x + 1)
        for x in range(self.n):
            if graph[x][1] == 1:
                to_visit = [(y, graph[x][1] + 1) for y in graph[x][0]]
                while to_visit:
                    visiting = to_visit.pop(-1)
                    graph[visiting[0] - 1][1] = max(graph[visiting[0] - 1][1], visiting[1])
                    for nxt in graph[visiting[0] - 1][0]:
                        to_visit.append((nxt, visiting[1] + 1))
        levels = [x[1] for x in graph]
        self.result = str(max(levels))

    def get_result(self):
        return self.result


if __name__ == "__main__":
    Solution = CodeforcesTask115ASolution()
    Solution.read_input()
    Solution.process_task()
    print(Solution.get_result())
