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 traceback
import sys

def solve(sup):
    tab = [-1]*len(sup)
    return max([to_root(sup,tab,i) for i in range(len(sup))])+1

def to_root(sup, tab, i):
    if tab[i] == -1:
        if sup[i] == -2:
            tab[i] = 0
        else:
            tab[i] = 1+to_root(sup,tab,sup[i])

    return tab[i]


def main():
    n = int(raw_input())
    sup=[]
    for i in range(n):
        x = int(raw_input())
        sup.append(x-1)        

    print solve(sup)

try:
    sys.setrecursionlimit(5000)
    main()
except Exception,e:
    print e
    traceback.print_exc(file=sys.stdout)

