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 *


class graph:
    # initialize graph
    def __init__(self, gdict=None):
        if gdict is None:
            gdict = defaultdict(list)
        self.gdict, self.edges = gdict, []

    # Get verticies
    def get_vertices(self):
        return list(self.gdict.keys())

    # add edge
    def add_edge(self, node1, node2):
        self.gdict[node1].append(node2)
        # self.gdict[node2].append(node1)
        self.edges.append([node1, node2])

    def bfs_util(self, i):
        queue, self.visit[i], level = deque([[i, 1]]), 1, 1

        while queue:
            # dequeue parent vertix
            s, level = queue.popleft()

            # enqueue child vertices
            for i in self.gdict[s]:
                if self.visit[i] == 0:
                    queue.append([i, level + 1])
                    self.visit[i] = 1
        return level

    def bfs(self):
        self.visit, self.cnt = defaultdict(int), 0

        for i in pars:
            if self.visit[i] == 0:
                self.cnt = max(self.bfs_util(i), self.cnt)
        return self.cnt


n, g, pars = int(input()), graph(), []
for i in range(1, n + 1):
    x = int(input())
    if x == -1:
        pars.append(i)
    else:
        g.add_edge(x, i)

print(g.bfs())

