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 

hs = {}
used = []
p = []
edges = []
n = 0

def rec(v, depth):

	st = []
	st.insert(0, (v, depth))

	while st:
		v, depth = st.pop(0)
		used[v] = True
		if depth in hs:
			hs[depth].append(v)
		else:
			hs[depth] = [v]

		for e in xrange(n):
			if edges[v][e] and not used[e]:
				st.insert(0, (e, depth + 1))
		depth -= 1

def solve():
	global p, used, edges, n

	n = int(raw_input())
	edges = [[False for _ in xrange(n)] for _ in xrange(n)]
	used = [False] * n
	p = [-1] * n
	for i in xrange(n):
		p[i] = int(raw_input())
		if p[i] != -1:
			p[i] -= 1
			edges[p[i]][i] = True

			# edges[i][p[i - 1]] = True
	for i in xrange(n):
		if not used[i] and p[i] == -1:
			rec(i, 1)
	print len(hs)

if __name__ == '__main__':
	solve()