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 math
import sys
from collections import defaultdict
from collections import deque
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
INF = 10 ** 18
sys.setrecursionlimit(1000000)
MOD = 1000000000 + 7
import heapq


first = 1 #if first node is 1
G = defaultdict(list)
def addEdge(a,b):
    G[a].append(b)
    G[b].append(a)
	
def BFSUtil(queue,vis,level):
    while queue:
        s = queue.popleft()

        for child in G.get(s,[]):
            if vis[child] == False:
                queue.append(child)
                level[child] = level[s] + 1
                vis[child] = True
            
def nodesatlevel(node,N,maxi,A):
    vis = [False for i in range(N+1)]
    
    for node in A:
        if vis[node] == False:
            level ={}
            queue = deque()
            level[node] = 0
            queue.append(node)
            vis[node] = True
            BFSUtil(queue,vis,level)
            maxi = max(maxi,max(level.values()))
    return maxi
	
N = int(input());maxi = -1;A = []
for i in range(N):
    u, = ilele()
    if u == -1:
        A.append(i+1)
        maxi = 0
    else:
        addEdge(i+1,u)
x = nodesatlevel(1,N,maxi,A)
#print(G,A)	
print(x+1)


