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 collections import defaultdict as dd
class Graph:
    def __init__(self,n):
        self.graph=dd(list)
        self.n=n
        self.vis=[False]*(n+1)
    def add_edge(self,u,v):
        self.graph[u].append(v)
    
    def bfs(self,u):
        q=[u]
        lvl=[0]*(self.n+1)
        lvl[u]=1
        while len(q)>0:
            f=q.pop()
            # print("I--->",f)
            for i in self.graph[f]:
                if self.vis[i]==False:
                    self.vis[i]=True
                    q.append(i)
                    lvl[i]=lvl[f]+1
        # print(lvl)
        return max(lvl)


n=int(input())
g=Graph(n)
l=[]
for i in range(n):
    x=int(input())
    if x==-1:
        l.append(i+1)
        continue
    g.add_edge(x,i+1)
# print(g.graph)
ans=-float("inf")
for i in l:
    k=g.bfs(i)
    # print(i,k)
    if k!=1:
        ans=max(ans,k)
if ans==-float("inf"):
    ans=1
print(ans)