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

class Graph(object):
    def __init__(self, n):
        self.root = []
        self.adj = {}
        self.make_adj(n)
    
    def make_adj(self, n):
        for i in range(1, n+1):
            self.adj[i] = []
    
    def adj_list(self, u):
        return self.adj[u]
class BFSResult:
	def __init__(self):
		self.level = {}
		self.parent = {}
		self.max_level = None
        
def bfs(g, s, r):
	r.parent[s] = None
	r.level[s] = 0	
	
	queue = deque()
	queue.append(s)
	while queue:
		u = queue.popleft()
		for n in g.adj_list(u):
			if n not in r.level:
				r.parent[n] = u
				r.level[n] = r.level[u] + 1
				r.max_level = r.level[u] + 1
				queue.append(n)
	#return r
if __name__ == '__main__':
    n = int(input().rstrip())
    g = Graph(n)
    i = 1
    r = BFSResult()
    for _ in range(n):
        u = int(input().rstrip())
        if u == -1:
            g.root.append(i)
        else:
            g.adj[u].append(i)
        i+=1
    #print(g.root)
    for s in g.root:
        bfs(g, s,r)
    max_l = 0
    for i in r.level:
        if r.level[i]>max_l:
            max_l = r.level[i]
    if max_l == 0:
        print(1)
    else:
        print(max_l+1)
    #print(r.level)
    
        
            
             

                
            
                
            
