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 os.path
from math import gcd, floor, ceil
from collections import *
import sys
mod = 1000000007
INF = float('inf')

def st(): return list(sys.stdin.readline().strip())
def li(): return list(map(int, sys.stdin.readline().split()))
def lis(): return list(map(str, sys.stdin.readline().split()))
def mp(): return map(int, sys.stdin.readline().split())
def inp(): return int(input())
def pr(n): return sys.stdout.write(str(n)+"\n")
def prl(n): return sys.stdout.write(str(n)+" ")

if os.path.exists('input.txt'):
    sys.stdin = open('input.txt', 'r')
    sys.stdout = open('output.txt', 'w')

mod = 1000000007
INF = float('inf')
Y = "YES"
N = "NO"

def bfs(g, i):
    q = [i]
    grp=0
    while len(q):
        n = len(q)
        while n:
            cur=q.pop(0)
            for nbr in g[cur]:
                q.append(nbr)
            n-=1
        grp+=1
    return grp


def solve():
    n = inp()
    g = [[] for _ in range(n+1)]
    for i in range(n):
        s = inp()
        if s==-1:
            continue
        g[s].append(i+1)

    grps = float("-inf")
    for i in range(1, n+1):
        grps = max(grps, bfs(g, i))

    print(grps)            

for test in range(1):
    solve()
