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 
def helper(employer, cur, dp):
    s = []
    while dp[cur] is None and employer[cur] > -1:

        s.append(cur)
        cur = employer[cur]

    if employer[cur] == -1:
        dp[cur] = 1
    top = dp[cur]
    while s:

        cur = s.pop()
        dp[cur] = top + 1
        top += 1




def ans():

    #inputs = list(map(int, input().strip().split()))
    employer = []
    while True:
        try:
            employer.append(int(input()))
        except EOFError:
            break


    if len(employer) == 0:
        return [0]
    n = employer[0]

    dp = [None] * (n + 1)
    dp[0] = 0
    s  = []
    for i in range(1, n + 1):
        if dp[i] is None:
            helper(employer, i, dp)
    return dp


print(max(ans()))
	 		   			      	 		 		 	  	 		