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 __future__ import division, print_function;import os;import sys
if sys.version_info[0] < 3:from __builtin__ import xrange as range;from future_builtins import ascii, filter, hex, map, oct, zip 
input = lambda: sys.stdin.readline().rstrip("\r\n");
ii = lambda: int(input());ll = lambda: list(map(int , input().split()));vv = lambda: map(int , input().split());pr = print;pl = lambda l: pr("".join(list(map(str,l))))
from collections import deque
sources = deque()
n = ii()
p = [0]*(n+1)
ans = p[:]
g = [[] for i in range(n+1)]
for i in range(1,n+1):
    x = ii()
    p[i] = x
    if x==-1:
        sources.append(i)
        ans[i] = 1
    else:
        g[x].append(i)
def BFS(): 
    visited = [False] * (n+1) 
    while sources: 
        s = sources.popleft() 
        visited[s] = True
        if p[s]!=-1:
            ans[s] = ans[p[s]]+1 
        for i in g[s]:
            if visited[i] == False: 
                sources.append(i)
BFS()
mp = {}
answer = 0
for i in range(1,n+1):
    if ans[i] in mp:
        mp[ans[i]].append(i)
    else:
        mp[ans[i]] = [i]
    answer = max(answer,ans[i])
print(answer)