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 bfs(adj,s):
    Q=[]
    Q.append(s)
    visited[s]=True
    level[s]=-1
    while len(Q)!=0:
        v=Q.pop(0)
        
        for ww in range(len(adj[v])):
            w=adj[v][ww]
            if visited[w]==False:
                visited[w]=True
                Q.append(w)
                level[w]=level[v]+1

n=int(input())
adj=[[] for _ in range(n+1)]

for i in range(1,n+1):
    u=int(input())
    if u==-1:
        adj[0].append(i)
    else:
        adj[u].append(i)
        
#print(adj)
            

level=[0]*(n+1)
visited=[False]*(n+1)
for i in range(0,n+1):
    if visited[i]==False:
        bfs(adj,i)
#print(level)
print(max(level)+1) 



