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 
# Understood the question

from collections import defaultdict
from collections import deque
# We have figured out the data structure to represent

class NTree:
    def __init__(self):
        self.hmap = defaultdict(list)

    def add_node(self, parent, node):
        if parent == -1:
            self.hmap[0].append(node)
        else:
            self.hmap[parent].append(node)

if __name__ == '__main__':
    n = int(input())
    t = NTree()
    i = 1
    while i <= n:
        t.add_node(int(input()), i)
        i += 1

    #print("The map is:", t.hmap)
    levels = 0
    node = t.hmap[0]
    q = deque([node])
    #print(q)

    level_children = []
    while q:
        #print("The queue is:", q)
        np = q.popleft()
        #print("The node is:", np)
        level_children = []
        for node in np:
            if node in t.hmap:
                for child in t.hmap[node]:
                    level_children.append(child)
        #print("The level children are:", level_children)
        if level_children:
            q.append(level_children)
        levels += 1

    print(levels)