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 queue import Queue

def heightByBfs(src, graph):
    q = Queue()

    q.put(src-1)

    currLevel = 1
    nxtLevel = 0
    level = 0

    while not q.empty():
        curr = q.get()

        currLevel-=1

        for child in graph[curr]:
            q.put(child-1)
            nxtLevel+=1
        

        if currLevel == 0:
            level+=1
            currLevel = nxtLevel
            nxtLevel = 0
    

    return level
    


def solve():
    #Form A Graph

    sources = []

    

    n = int(input())

    

    graph = [[] for _ in range(n)]

    for i in range(n):
        src = int(input())
        

        yeah = False


        if src == -1:
            sources.append(i+1)
        
        else:
            graph[src-1].append(i+1)
        

    
    if yeah: print("What is the damn BUG!!!!!!")

    res = 0


    for src in sources:
        res= max(res, heightByBfs(src, graph))

    
    print(res)

    return


solve()
    
