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 
from sys import stdin,stdout
def DFS_FOR_UNDIRECTED_GRPAH_USING_ADJACENCY_LIST(graph,root,depth):
    visited=[False]*len(graph)
    queue=[]
    queue.append(root)
    DFS=[]
    while(len(queue)!=0):
        v=queue.pop()
        if(visited[v]==False):
            DFS.append(v)
        if(visited[v]==False):
            visited[v]=True
            for i in range(0,len(graph[v])):
                queue.append(graph[v][i])
                depth[graph[v][i]]=depth[v]+1
    return DFS
n=int(stdin.readline())
emp=[]
minus1=[]
for i in range(0,n):
    emp.append([])
for i in range(0,n):
    chi=int(stdin.readline())
    if(chi!=-1):
        emp[chi-1].append(i)
    else:
        minus1.append(i)
#print(emp)
depth=[0]*(n)
m=0
for i in range(0,len(minus1)):
    DF=DFS_FOR_UNDIRECTED_GRPAH_USING_ADJACENCY_LIST(emp,minus1[i],depth)
    m=max(m,max(depth))
    depth=[0]*n
stdout.write(str(m+1)+'\n')
