human
Give me a Python program to print a single integer denoting the minimum number of groups that will be formed in the party given the following:
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 
The above is tricky. Write me a correct solution in PYTHON3.
assistant
#Here is Python program to print a single integer denoting the minimum number of groups that will be formed in the party.

from collections import *
class Graph():
    def __init__(self,n):
        self.n=n
        self.count=[0]*self.n
        self.g=defaultdict(list)
        self.visited=[False]*self.n
    def add_edge(self,u,v):
        self.g[v].append(u)
    def dfs(self,at,i):
        if(not self.visited[at]):
            self.count[i]+=1            
        for j in self.g[at]:
            self.dfs(j,i+1)
    def show(self):
        for i in range(self.n):
            print(i,self.g[i])
n=int(input())
graph=Graph(n+1)
for i in range(1,n+1):
    j=int(input())
    if(j==-1):
        graph.add_edge(i,0)
    else:
        graph.add_edge(i,j)  
graph.dfs(0,0)
res=0
for i in graph.count:
    if(i==0):
        break
    res+=1
print(res-1)