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 TreeNode:
    child = []
    val = -1

    def __init__(self, val):
        self.val = val
        self.child = []


def input_int():
    return int(raw_input()) 
 
 
def input_arr():
    return [int(x) for x in raw_input().split()] 
 
 
def printR(idx, itm):
    print("Case #"+str(idx)+": "+str(itm))

def printTree(node):
    print("node -->",node.val)
    child = ""
    for itm in node.child:
        print("child --> ",itm.val)

 
 
s = input_int()


node_map = {}
root_nodes = []
def update_node(node, parent):
    if node not in node_map:
        t1 = TreeNode(node)
        node_map[node] = t1

    t1 = node_map[node]

    if parent != -1:
        if parent not in node_map:
            t2 = TreeNode(parent)
            node_map[parent] = t2

        node_map[parent].child.append(t1)
        # printTree(node_map[parent])

    else:
        root_nodes.append(t1)



for i in range(s):
    itm = input_int()
    update_node(i+1, itm)
    # printTree(node_map[i+1])


# for key in node_map.keys():
#     printTree(node_map[key])


def max_depth(node_list):

    depth = 0
    while len(node_list) != 0:
        # print("node_list len -->",len(node_list))
        new_list = []
        for itm in node_list:
            # print("node. --> ", itm.val)

            new_list = new_list + itm.child

        depth += 1
        node_list = new_list

    return depth

print(max_depth(root_nodes))