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(raw_input())
nones = []

emp = []
for i in range(n):
    emp.append([])

for i in range(n):
    put = int(raw_input())
    if put != -1:
        emp[put - 1].append(i)
        
    else:
        nones.append(i)
memo = {}
mxdepth = 1
for i in nones:
    if emp[i] != []:
        arr = [(i,1)]
        ind = 0
        while ind != len(arr):
            #print emp, arr, ind
            for j in emp[arr[ind][0]]:
                if j not in arr:
                    depth = arr[ind][1] + 1
                    arr.append((j, depth))
                    if depth > mxdepth:
                        mxdepth = depth
            ind += 1

print mxdepth
