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 
class Node:
    def __init__(self, val):
        self.val = val
        self.depth = 0
        self.visited = False
        self.neighbors = []

def dfs(node):
    stack = []
    stack.append(node)
    while len(stack):
        root = stack.pop(-1)
        if(not root.visited):
            groups[root.depth] += 1
            root.visited = True
        for neighbor in root.neighbors:
            if(not neighbor.visited):
                neighbor.depth = root.depth+1
                stack.append(neighbor)

n = int(input())
groups = [0]*2001
nodes = [None]
for i in range(1, n+1):
    nodes.append(Node(i))
roots = []
for i in range(n):
    x =int(input())
    if(x == -1):
        roots.append(nodes[i+1])
        continue
    nodes[x].neighbors.append(nodes[i+1])


for node in roots:
    if(len(node.neighbors)):
        dfs(node)

count = 0
for group in groups:
    if(group):
        count += 1
if(count == 0):
    print(1)
else:
    print(count)