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

def main1():
    n = int(input())
    
    dict={}
    
    depth=[0]*(n+1)
    
    for i in range(1,n+1):
        readVal = int(input())
        
        if readVal!=-1:
            depth[i]=1
        
        if readVal in dict:
            dict[readVal].append(i)
            #depth[i]=1
        else:
            l=list()
            l.append(i)
            dict[readVal]=l
            
            
            
            
    #print(dict)
    #print(depth)
    
    def assignDepth(root):
        if root in dict:
            for child in dict[root]:
                depth[child]+=depth[root]
                assignDepth(child)
            
            #dict.pop(root)
                
                
    
    for x in dict:
        if x!=-1 and depth[x]==0:
            depth[x]=1
            #print(x)
            assignDepth(x)
                
    
    big=1
    
    for j in range(1,n+1):
        if depth[j]>big:
            big=depth[j]
            
    print (big)

if __name__=="__main__":
    sys.setrecursionlimit(5000)
    main1()
