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.Scanner;
import java.util.Vector;

public class Main {
	static Vector<Integer>[]v = new Vector[2005];
	static int Max = 0;
	static void dfs(int f, int cnt) {
		Max = Math.max(cnt, Max);
		//System.out.println(f);
		if(v[f].size() == 0) {
			return;	
		}else{
			for(int i = 0;i < v[f].size();i++) {
				++cnt;
				dfs(v[f].get(i), cnt);
				--cnt;
			}
		}
	}
	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		int n = cin.nextInt();
		for(int i = 1;i <= n;i++) {
			v[i] = new Vector<Integer>();
		}
		for(int i = 1;i <= n;i++) {
		
			int f = cin.nextInt();
			if(f != -1) {
				//	son[v] = i;
				v[f].add(i);
			}
		}
		for(int i = 1;i <= n;i++)
		dfs(i, 0);
		System.out.println(Max + 1);
	}
}

 		 	 	   	 	    		  	 	  	 		