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 
n =int(input())

gr=[]
st=[]
for i in range(n+1):
    gr.append([])

for i in range(n):

    x=int(input())

    if x!=-1:

        gr[x].append(i+1)

    else:
        st.append(i+1)

# print(gr)

def trav(a):

    dis = [-1 for x in range(n+1)]

    q=[]

    q.append(a)

    dis[a]=0

    while len(q)>0:

        b = q[0]
        q.pop(0)

        for i in range(len(gr[b])):

            z = gr[b][i]

            if dis[z]==-1:

                q.append(z)

                dis[z] = dis[b]+1

    dis.sort()

    return dis[len(dis)-1]
f=[]
for i in range(len(st)):

    k = trav(st[i])
    f.append(k)

f.sort(reverse = True)

print(f[0]+1)