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 
# 115A.py
# Party

n = int(raw_input())
g = [[] for _ in xrange(n+1)]
roots = []
m = 0
for i in xrange(n):
    a = int(raw_input())
    if a != -1:
        g[a].append(i+1)
    else:
        roots.append(i+1)

def dfs(G, v, d):
    max_depth = 1
    s = []
    s.append((v, 1))
    while s:
        vert = s.pop()
        max_depth = max(max_depth, vert[1])
        if d[vert[0]] == False:
            d[vert[0]] = True
            for w in G[vert[0]]:
                s.append((w, vert[1]+1))
            
    return max_depth

for i in roots:
    d = [False] * (n+1)
    m = max(m, dfs(g, i, d))

print m
