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 sys import stdin, stdout
import math,sys,heapq
from itertools import permutations, combinations
from collections import defaultdict,deque,OrderedDict
from os import path
import random
import bisect as bi
def yes():print('YES')
def no():print('NO')
if (path.exists('input.txt')): 
    #------------------Sublime--------------------------------------#
    sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
    def I():return (int(input()))
    def In():return(map(int,input().split()))
else:
    #------------------PYPY FAst I/o--------------------------------#
    def I():return (int(stdin.readline()))
    def In():return(map(int,stdin.readline().split()))
#sys.setrecursionlimit(1500)
def dict(a):
    d={} 
    for x in a:
        if d.get(x,-1)!=-1:
            d[x]+=1
        else:
            d[x]=1
    return d
def find_gt(a, x):
    'Find leftmost value greater than x'
    i = bi.bisect_left(a, x)
    if i != len(a):
        return i
    else:            
        return -1
def dfs(d,n,roots):
    visited=[False]*(n+1)
    parent=[0]*(n+1)
    ans=1
    for x in roots:
        q=deque([x])
        parent[x]=1
        while q:
            temp=q.popleft()
            for child in d[temp]:
                if visited[child]==False:
                    q.appendleft(child)
                    parent[child]=parent[temp]+1
                    ans=max(ans,parent[child])
            visited[temp]=True
    print(ans)
def main():
    try:
        n=I()
        roots=[]
        d=defaultdict(list)
        for x in range(n):
            q=I()
            if q==-1:
                roots.append(x+1)
            else:
                d[q].append(x+1)
        #print(d)
        #print(roots)
        dfs(d,n,roots)
    except:
        pass
        
M = 998244353
P = 1000000007
 
if __name__ == '__main__':
    #for _ in range(I()):main()
    for _ in range(1):main()