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.io.*;
import java.util.*;
import java.math.*;

public class Solution{
	public static void main(String[] args){
		Scanner scan = new Scanner(System.in);
	    int n = scan.nextInt();
	    
	    ArrayList<Integer>[] adjlist = new ArrayList[n];
	    for(int i=0;i<n;i++)
	        adjlist[i] = new ArrayList<Integer>();
	        
	    Queue<Integer> queue = new LinkedList<Integer>();
	    int[] best = new int[n];
	    Arrays.fill(best , -1);
	    
	    for(int i=0;i<n;i++){
	        int temp = scan.nextInt();
	        if(temp == -1){
	            queue.offer(i);
	            best[i] = 1;
	        }
	        else{
	            adjlist[temp-1].add(i);
	        }
	    }
	    
	    while(!queue.isEmpty()){
	        int top = queue.poll();
	        for(int i=0;i<adjlist[top].size() ; i++){
	            queue.offer(adjlist[top].get(i));
	            best[adjlist[top].get(i)] = best[top] + 1;
	        }
	    }
	    int res=0;
	    for(int i=0;i<n;i++)
	        res = Math.max(res,best[i]);
	        
	   System.out.println(res);
	}
}