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 

import java.util.*;

public class test
{
public static void dfs(int p, ArrayList list, int[] visited,int maxprov,ArrayList collect){
        if(list.get(p).equals(-1)){
            if(maxprov>(int)collect.get(collect.size()-1)){
                collect.add(maxprov);
            }
            maxprov=0;
        }
        else{
            visited[p]=1;
            maxprov++;
            int s=(int)list.get(p);
            dfs(s,list,visited,maxprov,collect);  
        }
        
}

public static void main (String[] args) {
        Scanner sc = new Scanner(System.in); 
        ArrayList<Integer> collect = new ArrayList<Integer>();
        collect.add(1);
        int maxprov=1;
        int[] visited;
        int k = sc.nextInt();
        ArrayList<Integer> list = new ArrayList<Integer>(); 
        list.add(-2);
       visited = new int[k+1];
       visited[0]=0;
       for( int j=1; j<k+1 ; j++){
            visited[j]=0;
        }
        for( int i=0; i<k ; i++){
            list.add(sc.nextInt());
        }
        for(int p=1 ;p<k+1; p++){
            if(visited[p]==0){
                test.dfs(p,list,visited,maxprov,collect);
            }
        }
    System.out.println(collect.get(collect.size()-1));
    }   
}


