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.LinkedList;
import java.util.Scanner;

public class A115_Party {
	
	private static int n;
	private static LinkedList [] adjList;
	private static boolean [] roots;
	private static int maxDepth; 
	
	static Scanner scan = new Scanner(System.in);
	
	public static void main(String[] args) {
	
		n = scan.nextInt();
		adjList = new LinkedList[n];
		roots = new boolean [n];
		
		for (int i = 0; i < n; i++) {
			adjList[i] = new LinkedList<>();
		}
		
		for (int i = 0; i < n ; i++) {
			int temp = scan.nextInt();
			if (temp != -1) {
				adjList[temp-1].add(i);
				
			} else {
				roots[i] = true;
			}
		}
		functionTrees ();
		System.out.println(maxDepth);
	}
	
	private static void functionTrees() {
		
		for (int i = 0 ; i < n; i++) {
			if (roots[i]) {
				DFS(i,0);
			}
		}
		
	}
	
	private static void DFS (int i, int count) {
		
		count++;
		if (count > maxDepth) {
			maxDepth = count;
		}
		int len = adjList[i].size();
		for (int j = 0; j < len; j++) {
			DFS ((Integer) adjList[i].get(j), count);
			//count--;
		}
	}
}
