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 
#Party3.py
n = int(input())
arr = [[0 for _ in range(n)] for _ in range(n)]
roots = []
for i in range(n):
    x = int(input())
    if x == -1:
        arr[i][i] = -1
        roots.append(i)
    else:
        arr[i][x - 1] = 1

longest = -1
#print(roots)
#print(arr)
for x in roots:
    v = [False] * n
    stack = []
    l = 0
    stack.append(x)
    while True:
        count = len(stack)
        if count == 0:
            break
        l += 1
        while count > 0:
            e = stack.pop(0)
            v[e] = True
            #print('parent: ', e)
            for i in range(n):
                if arr[i][e] == 1 and not v[i]:
                    stack.append(i)
                    #print('child: ', i)
            count -= 1
        #print('level: ', l)
    if l > longest:
        longest = l

print(longest)