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 
#                          JAI SHREE RAM

import math; from collections import *
import sys; from functools import reduce

def get_ints(): return map(int, input().strip().split())
def get_list(): return list(get_ints())
def getlistofstring(): return list(input().strip().split())
def printxsp(*args): return print(*args, end="")
def printsp(*args): return print(*args, end=" ")

UGLYMOD = int(1e9)+7; SEXYMOD = 998244353; MAXN = int(1e5)

# sys.setrecursionlimit(10**6)
# sys.stdin=open("input.txt","r");sys.stdout=open("output.txt","w")

# for _testcases_ in range(int(input())):
N = int(input())
manager = defaultdict(list)
queue = deque()
depth = [0] * (N+1)
for sub in range(1, N+1):
    man = int(input())
    if man == -1:
        queue.append([sub, 1])
        depth[sub] = 1
    else:
        manager[man].append(sub)
# print(queue)
# print(manager)
while len(queue):
    # print(queue)
    this = queue.popleft()
    man = this[0]
    dep = this[1] + 1
    for sub in manager[man]:
        if depth[sub] != dep:
            queue.append([sub, dep])
            depth[sub] = dep
print(max(depth))
        






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