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 math
from collections import deque

def findDepth(val,dic):
    q=deque()
    d=deque()
    c=val
    q.append(c)
    d.append(1)
    dp=[]
    while len(q)>0 :
        c=q[0]
        dHere=d[0]
        dp.append(dHere)
        q.popleft()
        d.popleft()
        for k,v in dic.items():
            if v==c:
                q.append(k)
                d.append(dHere+1)
    return max(dp)
        # if ~c.isLeaf():
        #     loopBoolean=True
        #     cc=c.firstChild
        #     q.append(cc)
        #     while loopBoolean:
        #         if cc.nextNode!=None:
        #             cc=cc.nextNode
        #             q.append(cc)
        #         else:
        #             loopBoolean=False

n=int(input())
dic1={}

for i in range(n):
    p2=int(input())
    p1=i+1
    dic1[p1]=p2
    if p2==-1:
        dic1[p1]="ROOT"
    if (p2 not in dic1) and (p2!=-1):
        dic1[p2]="ROOT"

# print(dic1)
depth=[]
for key,value in dic1.items():
    if value=="ROOT":
        depth.append(findDepth(key,dic1))
print(max(depth))