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.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.*;




public class B {
	static ArrayList<Integer>[]arr;
	
	public static void main(String[] args) throws IOException {
		Scanner sc=new Scanner (System.in);
		int n=sc.nextInt();
		ArrayList<Integer>arr[]=new ArrayList[n+1];
		for(int i=0;i<=n;i++)arr[i]=new ArrayList<>();
		Stack<Integer>stack=new Stack<>();
		int res=1;
		int vis[]=new int[n+1];
		for(int i=1;i<=n;i++) {
			int b=sc.nextInt();
			if(b==-1) {
				stack.push(i);
				vis[i]=1;
			}else {
				arr[b].add(i);
				
				
			}
			
			
			
		}
		while(!stack.isEmpty()) {
			int u=stack.pop();
			for(int v:arr[u]) {
				stack.push(v);
				vis[v]=vis[u]+1;
				res=Math.max(res, vis[v]);
			}
		}
		System.out.println(res);
	}

		
		
		
	
}