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 
from collections import defaultdict

class Graph():

        def __init__( self, n ):
                self.nodes = n
                self.graph = defaultdict( list )
                self.visited = [0] * self.nodes
        
        def add_edge( self, x, y ):
                self.graph[ x ].append( y )

        def max_length( self ):
                ret = 0
                for i in range( self.nodes ):
                        if self.visited[i] == 0:
                               ret = max( ret, self.dfs( i ) )
                return ret
        
        def dfs( self, u ):
                for v in self.graph[ u ]:
                        if self.visited[ v ] == 0:
                                self.dfs( v )
                        self.visited[ u ] = max( self.visited[ u ], self.visited[ v ] )
                self.visited[ u ] += 1
                return self.visited[ u ]

if __name__ == "__main__":
        
        n = int( input() )

        graph = Graph( n )

        for i in range( n ):
                p_i = int( input() )
                if not p_i == -1:
                        graph.add_edge( p_i - 1, i )

        print( graph.max_length() )

