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 
n = int(input())
l1=[]
i=0
while i<n:
    l1.append([int(input())])
    i+=1
i=0
while i<len(l1):
    targ=l1[i][0]
    while targ!=-1:
        l1[i].append(targ)
        l1[targ-1].append(i+1)
        targ=l1[targ-1][0]
    i+=1

i=0
tables=[[]]

while i<n:
    j=0
    while j<len(tables):
        ruined=0
        for x in tables[j]:
            if x in l1[i]:
                ruined=1
                break
        if ruined:
            j+=1
        else:
            tables[j].append(i+1)
            break
    if j==len(tables):
        tables.append([i+1])
    i+=1
print(len(tables))
