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 numpy as np 
N = int( input() )
ls = []
ns = []
for i in range( N ):
    ls.append( int(input()) - 1  )
    # print( ls[-1])
    ns.append( [] )
try:
    for i in range( N ):
        if ls[i] < 0 :
            continue
        ns[ls[i]].append( i )
except:
    print( 0 )
# print( ns )
try:
    deps = [-1] * N 
    def consTree( root, dep ):
        stacks = [(root,dep)]
        while stacks:
            root,dep = stacks.pop( )   
            deps[root] = dep 
            for i in ns[root]:
                stacks.append( (i,dep+1) )
except:
    print( 1 )
try:

    for i in range( N ):
        if ls[i] < 0:
            consTree( i, dep = 1 )
except:
    print( 2 )
print( max(deps) )

