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
import atexit
import io
import sys
 
# input = sys.stdin.readline
_OUTPUT_BUFFER = io.StringIO()
sys.stdout = _OUTPUT_BUFFER
 
@atexit.register
def write():
    sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
 
def array_to_string(a):
    s = ""
    for i in a:
        s += str(i)
    return s

def gcd(a,b): 
    if a == 0: 
        return b 
    return gcd(b % a, a) 
  
def lcm(a,b): 
    return (a*b) / gcd(a,b) 
 
# t = int(input())
# while t>0:
#     t = t-1
n = int(input())
d = dict()
depth = dict()
# for i in range(1,n+1):
#     depth[i] = 0
temp = []
for i in range(n):
    inp = int(input())
    if inp != -1:
        if depth.get(inp):
            depth[i+1] = depth[inp]+1
        else:
            temp.append((inp,i+1))
    else:
        depth[i+1] = 1
while len(temp)>0:
    for i in temp:
        if depth.get(i[0]):
            depth[i[1]] = depth[i[0]]+1
            temp.remove((i[0],i[1]))
# print(depth)
print(max(depth.values()))