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 
import sys
sys.setrecursionlimit(2050)
n=int(input())
d={}
ll=[]
mx=0
roots=[]
for i in range(1,n+1):
    d[i]=set()
for i in range(1,n+1):
    m=int(input())
    
    if m==-1:
        roots.append(i)
    else:
        #
        d[m].add(i)
 
 
level=[-1]*(n+1)  
lvl=1
dep=0
ans=-1
def dfs(s):
    global visited,n,dep,ans,K
    dep+=1
    if dep>ans:
        #print(K,s,dep)
        ans=dep
    #print(K,s,dep)
    visited[s]=1
    
    for i in d[s]:
        if visited[i]==0:
            dfs(i)
    dep-=1
 
for K in roots:
    dep=0
    visited=[0]*(n+1)
    dfs(K)
    for j in range(1,n+1):
        if visited[j]==0:
            #print('ans is',ans)
            if dep>=ans:
                ans=dep
            
            dfs(j)
        break
    
        
        
print(ans)