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 
n = int(input())

boss = []

super_b = []

for i in range(n):
	boss.append([])

for i in range(n):
	b = int(input())
	if b!=-1:
		boss[b-1].append(i)

	else:
		super_b.append(i)


# print(boss)
# print(super_b)
num = 0
for i in super_b:
	pres_boss = i
	level = 1
	queue = [pres_boss]
	marked = [0]*n
	marked[pres_boss] = 1
	# print(boss)
	while queue:
		pres_boss = queue[0]
		new = []

		for j in boss[pres_boss]:
			if marked[j]!=1:
				new.append(j)
				marked[j] = marked[pres_boss]+1
		# print(marked, pres_boss)
		# print(queue, new)
		if new:
			queue.extend(new)
		queue.pop(0)
		level = max(marked)
		

	if level>num:
		num = level

print(num)

