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 defaultdict, deque

class Tree(object):
    def __init__(self, root):
        self.root = root

    @property
    def longest_chain(self):
        """Compute chain with the longest length"""

        stack = deque()
        stack.appendleft([self.root, 1])

        max_depth = 0
        while stack:
            node, depth = stack.popleft()
            for child in node.children:
                stack.appendleft([child, depth+1])

            max_depth = max(depth, max_depth)

        return max_depth

class Node:
    def __init__(self, id):
        self.id = id
        self.children = []

    def __repr__(self):
        s = 'id:{}'.format(str(self.id))
        if self.children:
            s += '->{}'.format(self.children)

        return s


if __name__ == '__main__':
    n = int(raw_input())
    nums = [int(raw_input()) for _ in range(n)]

    # Compute children in reverse order. i.e. if A is a supervisor of B is
    # represented as A -> B.
    children, root_ids = defaultdict(list), []
    for id, num in enumerate(nums):
        if num == -1:
            root_ids.append(id)
            continue
        children[num-1].append(id) # zero-indexing

    nodes = {id: Node(id) for id in range(n)}
    for id, child_ids in children.items():
        nodes[id].children = [nodes[child_id] for child_id in child_ids]

    roots = [nodes[root_id] for root_id in root_ids]
    trees = [Tree(root) for root in roots]

    longest_chain = 0
    for tree in trees:
        longest_chain = max(tree.longest_chain, longest_chain)

    print longest_chain
