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 sys
from collections import deque
num_of_soldier = int(sys.stdin.readline())

visited_flag = [-1]*(num_of_soldier+1)
graph = [0]*(num_of_soldier+1)

for i in range(num_of_soldier):
    graph[i+1] = int(sys.stdin.readline())


def dfs(start_node):
    stack = deque()
    visited_node = []
    num=0
    stack.append(start_node)
    visited_node.append(start_node)
    while stack:
        node = stack.pop()
        if visited_flag[node] != -1:
            num+= visited_flag[node]
            break
        if graph[node]==-1:
            visited_flag[node] = 0
            break
        
        num+=1
        stack.append(graph[node])
        visited_node.append(graph[node])


    for i in visited_node:
        visited_flag[i] = num
        num-=1

for i in range(num_of_soldier):
    if visited_flag[i+1] == -1:
        dfs(i+1)
print(max(visited_flag)+1)
    
