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 
from queue import Queue
def minSet(n,ch,d,q):
    maxm = 1
    while not q.empty():
        u = q._get()
        for v in ch[u]:
            d[v] = d[u] + 1            
            q._put(v)
            if maxm < d[v]:
                maxm = d[v]
    return maxm

n = int( input() )
ch = []
d = []
q = Queue()
for i in range( 0 , n + 1 ):
    ch.append([]);
    d.append(0)
for i in range( 1 , n + 1 ):
    f = int( input() )
    if f == -1:
        d[i] = 1
        q._put(i)
    else:
        ch[f].append(i)
print(minSet(n,ch,d,q))
