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())
u = dict()
Root = list()
for i in range(0, n):
    Father = int(input())
    if Father == -1:
        Root.append(i+1)
        continue
    if Father not in u: u[Father] = list() 
    u[Father].append(i+1)
#print('u=',u)
maxH = 1
for i in Root:
    Q = list()
    Q.append(i)
    H = 1
    Wide = [None] * 100000
    Wide[0], Wide[1] = 0, 1
    while Q:
        t = Q.pop()
        if Wide[H]>0:
            Wide[H] = Wide[H] - 1
        else:
            while Wide[H] == 0:
                H = H-1
            Wide[H] = Wide[H] - 1
      #  print('t=', t)
      #  print('Q=', Q)
      #  print('Wide=', Wide[1:H+1])
      #  print('H=', H)
        if t not in u: continue
        H = H+1
        if H > maxH:
            maxH = int(H)
        Wide[H] = len(u[t])
        Q.extend(u[t])
print(maxH)
