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
class Graph:
    def __init__(self):
        self.graph = defaultdict(list)
    
    def addEdge(self,u,v):
        self.graph[u].append(v)

    def dfs(self,visited,v,num,group):
        visited[v] = True
        group[v] = num
        for i in self.graph[v]:
            if(visited[i]==False):
                self.dfs(visited,i,num+1,group)

    def DFSUtil(self,root,V):
        visited = [False]*(V+1)
        group = [-1]*(V+1)
        num = 1
        for i in range(1,V+1):
            if(root[i]==True):
                self.dfs(visited,i,num,group)
        return max(group)

n = int(input())
root = [True]*(n+1)
g = Graph()
for i in range(1,n+1):
    k = int(input())
    if(k!=-1):
        g.addEdge(k,i)
        root[i]=False
print(g.DFSUtil(root,n))