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 bisect
from collections import defaultdict
from collections import deque
import math

def ni():
    return int(raw_input())

def nis():
    return map(int, raw_input().split())

def si():
    return raw_input()

def sis():
    return raw_input().split()

def bfs(n, r, a, p):
    height = [0 for i in range(n + 1)]
    q = deque(r)
    while q:
        u = q.popleft()
        height[u] = height[p[u]] + 1
        for v in a[u]:
            q.append(v)

    return max(height)

n = ni()
a = [list() for i in range(n + 1)]
r = []
p = [-1]
for i in range(1, n + 1):
    val = ni()
    p.append(val)
    if val == -1:
        val = 0
        r.append(i)
    a[val].append(i)

print bfs(n, r, a, p)

