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 
# Still Incomplete
# Try 1+1
#graph, trees, dfs and similar
from collections import defaultdict
MaxList = []
class Graph:
    def __init__(self):
        self.graph = defaultdict(list)
    def addEdge(self,fr,to):
        self.graph[fr].append(to)
        self.graph[to].append(fr) #Don't Forget This
    def BFS(self,root,Visited):
        queue = []
        queue.append(root)
        Visited[root-1] = True
        count = 0
        while queue:
            InnerQueue = queue.copy()  #Copy Method Necessary
            while InnerQueue:       #One Breadth At A Time
                InnerQueue.pop(0)
                current = queue.pop(0)
                for i in self.graph[current]:
                    if(Visited[i-1]==False):
                        queue.append(i)
                        Visited[i-1] = True
            count+=1                #Counting Height Here
        MaxList.append(count)

N = int(input())
G = Graph()
Visited = [False] * N
List = []
for i in range(N):
    temp = int(input())
    if(temp != -1):
        G.graph[temp].append(i+1)
    else:
        List.append(i+1)
for i in List:
    G.BFS(i,Visited)
print(max(MaxList))
