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 GetParentHt(hts,kid,parents):
    return hts[parents[kid-1]-1]+1

def GetHts(roots,graph,parents):
    hts = [-1]*(len(graph))
    #print (len(roots))
    #print (parents)
    for root in roots:
        #print (root)
        hts[root-1] = 0
        ht = 0
        stack = [root]
        while stack:
            curr =stack.pop()
            for kid in graph[curr-1]:
                hts[kid-1] = GetParentHt(hts,kid,parents)
                stack.append(kid)
            #print (stack)
    return hts

def Party(hts):
    hts.sort()
    curr = hts[0]
    if hts:
        groups = 1
    else:
        groups = 0
    #print (hts)
    for i in hts:
        if i != curr:
            curr = i
            groups += 1

    return groups

def main():
    n = int(input())
    graph = {}
    for i in range(n):
        graph[i] = []
    #print (graph)
    roots = []
    parents = [0]*n

    for i in range(n):
        p = int(input())
        if p == -1:
            roots.append(i+1)
            parents[i] = -1
        else:
            graph[p-1].append(i+1)
            parents[i] = p
        #print (p,graph[p-1])
    hts = GetHts(roots,graph,parents)
    print (Party(hts))

main()
