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 
#!python3

from collections import deque, Counter
import array
from itertools import combinations, permutations
from math import sqrt
import unittest


def read_int():
    return int(input().strip())


def read_int_array():
    return [int(i) for i in input().strip().split(' ')]

######################################################

n = read_int()
P = [int(input().strip()) for _ in range(n)]
H = [0] * n
max_h = 0
for i in range(n):
    v = P[i]
    h = 1
    while v != -1 and H[v-1] < h:
        H[v-1] = h
        v = P[v-1]
        h += 1
    h -= 1
    if h > max_h:
        max_h = h
print(max_h+1)

"""
-1 1 2 1 -1
2 1 0 0 0
1

0  -1 1
1  1 1  -1 2
2  2 1  1 2  -1 3
3  1 1
4  -1 1
max_h = 2



"""

