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 
impor
What is 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.
t 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,parents=[]):
        #print(v,end=" ")
        visited[v]=True
        for i in self.adj_list[v]:
            if visited[i]==False:
                self.dfsUTIL(i,visited,parents)
                parents.append(i)
 
        
        
                
    
    def dfs(self,v):
        visited=[False]*(max(self.adj_list)+1)
        depth=0
        parents=[v]
        self.dfsUTIL(v,visited,parents)
        return len(parents)
        
        
        
#####sorting a dictionary by the values#####
def dict_sort(ans):
    ans=sorted(ans.items(),reverse=True,key=lambda kv:(kv[1]))
##### naive method for testing prime or not O(n^.5)#####
def is_prime(n):
    if n==1:
        return 0
    for i in range(2,int(math.sqrt(n))+1):
        if n%i==0:
            return False
    return True
#####swap function#####
def swap(a,b):
    temp=a
    a=b
    b=temp
    return a,b
#####Primes till Nth O(n)#####
def seive_primes(n):
    flag=[0]*(n+10)
    flag[1]=flag[0]=1
    i=2
    while i*i<=n+1:
        if flag[i]==0:
            j=i*i
            while j<=n+1:
                flag[j]=1
                j+=i
        i+=1
    return flag
#####greatest common divisor of two numbers#####
def gcd(a,b):
    if b==0:
        return a
    return gcd(b,a%b)
#####least common multiplyer of two...