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.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.util.Vector;


public class Main {
	
	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		Queue<Integer>q = new LinkedList<Integer>();
		Vector<Integer>[]e = new Vector[2005];
		int n = cin.nextInt();
		
		for (int i = 1; i <= n; i++) {
			e[i] = new Vector<Integer>();
		}
		
		for(int i = 1;i <= n;i++) {
			int v = cin.nextInt();
			if(v != -1)
			e[v].add(i);
		}
		int vis[] = new int[2005];
		
		int deep[] = new int[2005];
		int maxdp = 0, maxdp1 = 0;
		int ans = 0;
		for(int i = 1;i <= n;i++) {
			maxdp1 = 0;
			if(e[i].size() != 0 && vis[i] == 0) {
				
				q.offer(i);///i 作为根节点开始
				
				while(!q.isEmpty()) {
					int u = q.poll();
					maxdp = 0;
					for(int j = 0;j < e[u].size();j++) {
						int v = e[u].get(j);

						deep[v] = deep[u] + 1;
						if(vis[v] != 0) {///maxdp 是 分支的最大层数
							maxdp = Math.max(maxdp, vis[v] + deep[v] - 1);/// u - d
							continue;
						}
						vis[v] = 1;
						q.add(v);
						maxdp = Math.max(deep[v], maxdp);
						
					}
					
					maxdp1 = Math.max(maxdp, maxdp1);
				}
			}	
			vis[i] = maxdp1 + 1;
			ans = Math.max(vis[i], ans);
		}
		
		System.out.println(ans);
		
	}
}
