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 
class Graph:
	def __init__(self,v):
		self.v = v
		self.graph = [[0 for col in range(v)] for row in range(v)]
		self.visited = [False for _ in range(v)]

DEBUG = False
def getDepth(g,src):
	q = [src]
	depth = 0
	cnt = 1
	while q:
		val = q.pop(0)
		if DEBUG:
			print("==>" + str(val))
		cnt -= 1
		g.visited[val] = True
		for i in range(g.v):
			if g.graph[val][i]==1:
				if g.visited[i]==False:
					q.append(i)
		
		if cnt == 0:
			depth += 1
			cnt = len(q)
	return depth

def solve():
	v = int(input())
	g = Graph(v)
	root = []
	for i in range(v):
		parent = int(input())
		if parent == -1:
			root.append(i)
			continue
		g.graph[parent-1][i] = 1
	max = -1
	if DEBUG:
		print(g.graph)
	for i in range(len(root)):
		temp = getDepth(g,root[i])
		if DEBUG:
			print("======")
		if temp > max:
			max = temp
	
	print(str(max))
if __name__ == "__main__":
	solve()