human
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 
 Write me a Python code to print a single integer denoting the minimum number of groups that will be formed in the party?
assistant
#Here is Python code to print a single integer denoting the minimum number of groups that will be formed in the party.

n=int(raw_input())
dic={}
a=[0 for i in range(n+1)]
for i in range(1,n+1):
    dic[i]=[]
for i in range(1,n+1):
    p=int(raw_input())
    a[i]=p   
    if p !=-1:
        dic[p].append(i)
class node:
    def __init__(self,val,high):
        self.val=val
        self.high=high
q=[]
a2=[0 for i in range(n+1)]
for i in range(1,n+1):
   if a[i]==-1:
       q.append(node(i,1))
       a2[i]=1
z=1
while len(q)>0:
    top=q[0]
    del q[0]
    for j in dic[top.val]:
        if a2[j]==0:
            q.append(node(j,top.high+1))
            if top.high+1>z:
                z=top.high+1
            a2[j]=1
print z