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 main():
    from sys import stdin
    from collections import deque
    n = int(input())
    A = list(map(int,stdin.read().strip().split()))
    class g:
        def __init__(self):
            self.depth=-1
            self.child = []
    d = [g() for i in range(n)]
    
    M = []
    for i in range(n):
        if A[i] == -1:
            M.append(i)
            continue
        d[A[i]-1].child.append(i)
            
    def bfs(a):
        q = deque()
        q.append(a)
        a.depth=0
        while(len(q)>0):
            a = q.popleft()
            for i in a.child:
                if d[i].depth==-1:
                    d[i].depth=a.depth+1
                    q.append(d[i])
        
    for i in M:
        bfs(d[i])
    D=[]
    for i in range(n):
        D.append(d[i].depth)
        # print(d[i].child)
    print(max(D)+1)
    # print(D)
main()