human
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
assistant
#Here is Python code to print a single integer denoting the minimum number of groups that will be formed in the party.

import sys
sys.setrecursionlimit(100000)

class Graph:
    def __init__(self, n):
        self.n = n
        self.adj = [[] for i in range(n)]
    def add(self, v, u):
        self.adj[v].append(u)
        self.adj[u].append(v)

n = int(input())
g = Graph(n)
root = []
for i in range(n):
    p = int(input())
    if p == -1:
        root.append(i)
        continue
    g.add(i, p - 1)

used = [False for i in range(n)]
dp = [0 for i in range(n)]

def dfs(v, p, depth):
    used[v] = True
    dp[v] = depth
    for u in g.adj[v]:
        if used[u]:
            continue
        dfs(u, v, depth + 1)
for x in root:
    dfs(x, -1, 0)

print(max(dp) + 1)