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
sys.setrecursionlimit(10**6)
MOD = 1000000007
def get_array(): return list(map(int, sys.stdin.readline().split()))
def get_ints(): return map(int, sys.stdin.readline().split())
def input(): return sys.stdin.readline()
def print_array(a): print(" ".join(map(str, a)))
visited = [False] * 2005
dist = [1] * 2005
graph = [[] for i in range(2005)]
root = []
def explore(node):
    visited[node] = True
    for child in graph[node]:
        if visited[child] == False:
            dist[child] = 1 + dist[node]
            explore(child)
def dfs():
    for i in root:
        if visited[i] == False: explore(i)
def main():
    n = int(input())
    for i in range(1, n + 1):
        x = int(input())
        if x != -1: graph[x].append(i)
        else: root.append(i)
    dfs()
    print(max(dist[1:n + 1]))
if __name__ == "__main__":
    main()