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 
import sys
from collections import defaultdict
from time import time


class Node:
    def __init__(self, i, children=None):
        self.i = i
        self.children = children if children else []

    def __repr__(self):
        return "{}".format(self.i)


def find_max_level(root):
    stack = [(root, 1)]

    max_level = -1
    while stack:
        node, lvl = stack.pop()
        max_level = max(lvl, max_level)

        for child in node.children:
            stack.append((child, lvl + 1))

    return max_level


def main():
    data = sys.stdin.read()
    lines = data.split("\n")
    n_employees = int(lines[0])

    j = 1
    nodes = [ Node(i) for i in range(1, n_employees + 1) ]

    trees = []
    for i in range(1, n_employees + 1):
        mgr = int(lines[i])
        if mgr == -1:
            trees.append(nodes[i - 1])
        else:
            nodes[mgr  - 1].children.append(nodes[i - 1])

    # s = time()
    print(max(find_max_level(root) for root in trees))
    # e = time()
    # if (e - s) > 3:
    #     print('Time limit exceeded: {:.0f}'.format((e - s)))


if __name__ == "__main__":
    main()