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 
The above is tricky. Write me a correct solution in PYTHON.
assistant
#Here is Python code to print a single integer denoting the minimum number of groups that will be formed in the party.

"""Template for Python Competitive Programmers prepared by Mayank Chaudhary aka chaudhary_19"""

# to use the print and division function of Python3
from __future__ import division, print_function

"""value of mod"""
MOD = 998244353

"""use resource"""
# import resource
# resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])

"""for factorial"""
# fact = [1]
# for i in range(1, 100005):
#     fact.append((fact[-1] * i) % MOD)
# ifact = [0] * 100005
# ifact[100004] = pow(fact[100004], MOD - 2, MOD)
# for i in range(100004, 0, -1):
#     ifact[i - 1] = (i * ifact[i]) % MOD

"""uncomment next 4 lines while doing recursion based question"""
# import threading
# threading.stack_size(2**27)
import sys

# sys.setrecursionlimit(10**6)


"""uncomment modules according to your need"""
# from bisect import bisect_left, bisect_right, insort
# import itertools
# from math import floor, ceil, sqrt
# import heapq
# from random import randint as rn
# from Queue import Queue as Q
# from collections import Counter, defaultdict

'''
def modinv(n, p):
    return pow(n, p - 2, p)
'''

"""
def ncr(n, r, p):  # for using this uncomment the lines calculating fact and ifact
    t = ((fact[n]) * ((ifact[r] * ifact[n - r]) % p)) % p
    return t
"""

def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()

# def GCD(x, y):
#     while (y):
#         x, y = y, x % y
#     return x
#
# def lcm(x, y):
#     return (x*y)//(GCD(x, y))
#
# def get_xor(n):
#     return [n,1,n+1,0][n%4]

"""*******************************************************"""

def dfs(node, level, mydict, visited):
    visited[node] = True
    maxi = level
    for child in mydict[node]:
        if not visited[child]:
            maxi = max(maxi, dfs(child, level + 1, mydict, visited))
    return maxi

def main():

    n = int(input())
    mydict = {i : [] for i in range(1, n+1)}
    parent = {i : -1...