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 party(n, managers):
    # to find the longest DFS path
    def DFS(p):
        if p in memo:
            return memo[p]
        else:
            if managers[p] < 0:
                memo[p] = 1 # himself
                return 1
            else:
                res = 1 + DFS(managers[p])
                memo[p] = res
                return res

    memo = {}
    longest = 0
    for p in range(n):
        val = 0
        while p >= 0:
            val += 1
            p = managers[p]
        longest = max(longest, val)

    return longest

if __name__ == "__main__":  
    n = int(input()) # number of people

    managers = []    # store all manager numbers
    for i in range(n):
        manager = int(input()) - 1     # convert 1-index to 0-index
        managers.append(manager)
    
    res = party(n, managers)
    print(res)