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 
#!/usr/bin/python

n = input() 
parent = [-1] * (n+1)
dist = [0] * (n+1)

def findR(x) : 
    # if parent[x] < 0 : 
    #   return x
    # else :
    #   pre = parent[x]
    #   parent[x] = findR(parent[x])
    #   dist[x] = dist[pre] + dist[x]
    #   return parent[x]
    cur = x
    stack = []
    while True : 
        if parent[cur] < 0 : 
            break
        stack.append(cur)
        cur = parent[cur]
    
    while len(stack) > 0 : 
        top = stack[-1]
        pre = parent[top]
        dist[top] = dist[top] + dist[pre]
        parent[top] = cur
        stack.pop()

    return cur

def merge(x, y) : 
    # merge y to x
    rx = findR(x)
    ry = findR(y) 
    if rx == ry : 
        return 
    parent[rx] += parent[ry]
    parent[ry] = rx
    dist[ry] = dist[x] + 1

for i in range(1, n+1) : 
    node = input()
    if node == -1 :
        pass 
    else : 
        merge(node, i)

for i in range(1, n+1) : 
    findR(i)

layer = 0
cnt = 0
while True : 
    if any([dist[i] == layer for i in range(1, n+1)]) : 
        cnt += 1
        layer += 1
    else : 
        break

print cnt
