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 

Can you provide a Python code to print a single integer denoting the minimum number of groups that will be formed in the party?
assistant
#Here is Python code to print a single integer denoting the minimum number of groups that will be formed in the party.

from collections import defaultdict
import sys

sys.setrecursionlimit(3000)

class Tree:
    def __init__(self, data):
        self.tree = defaultdict(list)

        for node, parent in enumerate(data, start=1):
            self.tree[parent].append(node)

    def get_depth(self):
        return self.DFS(-1)

    def DFS(self, node, depth=0):
        m = 0
        for child in self.tree[node]:
            m = max(m, self.DFS(child, depth+1))

        return max(m, depth)

n = int(input())
p = [int(input()) for _ in range(n)]

t = Tree(p)
print(t.get_depth())