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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;

public class Main {
      static ArrayList <Integer> [] adjList; 
      static int color []; 
      static int parent[]; 
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in)); 
		int n = Integer.parseInt(bf.readLine()); 
		adjList = new ArrayList [n+1]; 
		parent = new int [n+1]; 
		 color = new int [n+1]; 
		for (int i = 0; i < adjList.length; i++) {
			adjList [i] = new ArrayList(); 
		}
		for(int i =1; i <=n ;i++) {
			int p = Integer.parseInt(bf.readLine()); 
			if (p!= -1)
			adjList[p].add(i); 
			parent[i] = p; 
		} 
		for (int i = 1; i <= n; i++) {
			if (parent[i]==-1)
				bfs(i); 
		}
		int max = -1;
		for (int i = 1; i <= n; i++) {
			max = Math.max(max, color[i]); 
		} 
		//9System.out.println(Arrays.toString(color));
		System.out.println(max+1);
		
 
	}
         static void bfs (int u ) {
           Queue <Integer> q = new LinkedList<Integer> (); 
           q.add(u); 
           while (!q.isEmpty()) {
        	   int parent = q.poll(); 
        	   for (int child: adjList[parent] ) {
        		   color[child] = 1 + color[parent]; 
        		   q.add(child); 
        	   }
           }
       }
}