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 
import queue

class nod:
    state = 0
    emp = []
    depth = 0

    def __init__(self, state, lis, depth):
        self.state = state
        self.emp = lis
        self.depth = depth

def bfs(G, source):
    Q = queue.Queue()
    level = [10000]*10000
    Q.put(source)
    level[source] = 0
    maxx = 0

    while Q.empty() == False:
        u = Q.get()

        if u in G:
            for edge in G[u]:
                if level[edge] == 10000:
                    level[edge] = level[u] + 1
                    maxx = max(level[edge], maxx)
                    Q.put(edge)

    return maxx + 1



hm = int(input())
sources = []
diction = {}
maxx = 1

for i in range(1, hm + 1):
    boss = int(input())
    if boss == -1:
        sources.append(i)
        pass
    else:
        if boss not in diction:
            diction[boss] = []
            diction[boss].append(i)
        else:
            diction[boss].append(i)

maximum = 0

for head in sources:
    maximum = max(maximum, bfs(diction, head))

print(maximum)
