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;readline = sys.stdin.buffer.readline
def i1(): return int(readline())
def nl(): return [int(s) for s in readline().split()]
def nn(n): return [int(readline()) for i in range(n)]
def nnp(n,x): return [int(readline())+x for i in range(n)]
def nmp(n,x): return (int(readline())+x for i in range(n))
def nlp(x): return [int(s)+x for s in readline().split()]
def nll(n): return [[int(s) for s in readline().split()] for i in range(n)]
def mll(n): return ([int(s) for s in readline().split()] for i in range(n))
def s1(): return readline().rstrip()
def sn(n): return [readline().rstrip() for i in range(n)]
def sm(n): return (readline().rstrip() for i in range(n))
def redir(s): global readline;import os;fn=sys.argv[0] + f'/../in-{s}.txt';readline = open(fn).readline if os.path.exists(fn) else readline

redir('a')
n = i1()
roots, children = [], [[] for i in range(n+1)]
# print(n)
for i in range(1,n+1):
    p = i1()
    if p >= 0:
        children[p].append(i)
    else:
        roots.append(i)
# print(roots, children)
maxdepth = 0
while roots:
    maxdepth += 1
    roots = [j for i in roots for j in children[i]]
    # for i in roots:
    #     print("i",i, children[i])
    #     for j in children[i]:
    #         print(j,end='')
    # print(maxdepth)
    # if maxdepth > 4: break
print(maxdepth)
