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 collections import Counter
from collections import defaultdict
import math
import random
import heapq as hq
from math import sqrt
import sys
from functools import reduce


def input():
    return sys.stdin.readline().strip()


def iinput():
    return int(input())


def tinput():
    return input().split()


def rinput():
    return map(int, tinput())


def rlinput():
    return list(rinput())


mod = int(1e9)+7


def factors(n):
    return set(reduce(list.__add__,
                      ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))


# ----------------------------------------------------

def bfs(g,n):
    depth = 0
    q=[0]
    visited=[False]*(n+1)
    visited[0]=True
    while len(q)>0:
        temp_queue=[]
        while len(q)>0:
            node=q.pop()
            for i in g[node]:
                if not visited[i]:
                    temp_queue.append(i)
                    visited[i]=True
        depth+=1
        q=temp_queue
    return depth-1
            

if __name__ == "__main__":
    num = iinput()
    g=defaultdict(list)
    for i in range(num):
        n=iinput()
        if n!=-1:
            g[n].append(i+1)
        else:
            g[0].append(i+1) #dummy variable
    print(bfs(g,num))
        
