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 
class Graph:
    def __init__(self, size):
        self.size = size
        self.adj_list = {}
        for i in range(size):
            self.adj_list[i] = []

    def add_edge(self, src, des):
        self.adj_list[src].append(des)

    def BFS(self, src):
        d2s = [-1 for i in range(self.size)]
        visited = [False for i in range(self.size)]
        d2s[src] = 0
        visited[src] = True
        queue = [src]

        while len(queue) != 0:
            u = queue.pop(0)
            for node in self.adj_list[u]:
                if not visited[node]:
                    visited[node] = True
                    d2s[node] = d2s[u] + 1
                    queue.append(node)
        return d2s


def main():
    n = int(input())
    g = Graph(n)
    roots=[]
    bfss=[]
    for i in range(n):
        u = int(input())
        if u != -1:
            g.add_edge(u - 1, i)
        else:
            roots.append(i)
    for i in roots:
        bfss.append(max(g.BFS(i)) + 1)
    print(max(bfss))


if __name__ == '__main__':
    main()
