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 
import math
import sys
##### graph implementation wth adjacancy list#####
class Graph:
    def __init__(self,Nodes,is_directed=False):
        self.nodes=Nodes
        self.adj_list={}
        self.is_directed=is_directed

        for node in self.nodes:
            self.adj_list[node]=[]
            
    def add_edge(self,u,v):
        if v not in self.adj_list[u]:
            self.adj_list[u].append(v)
        if self.is_directed==False:
            if u not in self.adj_list[v]:
                self.adj_list[v].append(u)

    def print_graph(self):
        for node in self.nodes:
            print(node,"->",self.adj_list[node])

    def degree_node(self,node):
        return len(self.adj_list[node])

    def dfsUTIL(self,v,visited):
        #print(v,end=" ")
        p=1
        visited[v]=True
        for i in self.adj_list[v]:
            if visited[i]==False:
                p=self.dfsUTIL(i,visited)+1
        return p
    def dfs(self,v):
        visited=[False]*(max(self.adj_list)+1)
        parents=self.dfsUTIL(v,visited)
        return parents

t=int(input())
nodes=[i+1 for i in range(t)]
graph=Graph(nodes,True)
for i in range(t):
    a=int(input())
    if a!=-1:
        graph.add_edge(i+1,a)
#graph.print_graph()
maxx=-1
for i in nodes:
    a=graph.dfs(i)
    if a>maxx:
        maxx=a
print(maxx)