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 
def __dfs(node, visited, graph, c):
    temp = c
    if(visited[node]):
        return c
    visited[node] = True
    for i in graph.get(node, []):
        c = max(c, __dfs(i, visited, graph, temp+1))
    return c


def dfsAll(n, graph):
    c = 1
    visited = [False]*n
    for ii in graph:
        if(visited[ii] == False):
            tt = __dfs(ii, visited, graph, 1)
            c = max(c, tt)
    print(c)


t = int(input())
graph = {}
tempg = []
for x in range(t):
    temp = int(input())
    tempg.append((x, temp))
tempg.sort(key=lambda x: x[-1])
for x in tempg:
    if(x[1] != -1):
        if(graph.get(x[1]-1) == None):
            graph[x[1]-1] = []
        graph[x[1]-1].append(x[0])
    elif(x[1] == -1):
        if(graph.get(x[0]) == None):
            graph[x[0]] = []
for i in range(t):
    if graph.get(i) == None:
        graph[i] = []
dfsAll(t, graph)
