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.Scanner;

public class Main {

	static ArrayList<Node> tree = new ArrayList<Node>();

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Node root = new Node();

		Scanner sc = new Scanner(System.in);
		int num = sc.nextInt();
		
		for (int i = 0; i < num; i++) {
			tree.add(new Node());
		}
		
		for (int i = 0; i < num; i++) {
			int tmp = sc.nextInt();
			if (tmp == -1) {
				tree.get(i).parentNode = root;
				root.childNode.add(tree.get(i));
			} else {
				tree.get(i).parentNode = tree.get(tmp - 1);
				tree.get(tmp - 1).childNode.add(tree.get(i));
			}
		}
		sc.close();

		System.out.println(findDeepth(root, 0));

	}

	static int findDeepth(Node root, int n) {
		if (root.childNode.size() == 0) {
			return n;
		} else {
			ArrayList<Integer> list = new ArrayList<Integer>();
			for (int i = 0; i < root.childNode.size(); i++) {
				list.add(findDeepth(root.childNode.get(i), n + 1));
			}
			int tmp = -1;
			for (int i = 0; i < list.size(); i++) {
				if (list.get(i) > tmp) {
					tmp = list.get(i);
				}
			}
			return tmp;
		}
	}

}

class Node {
	public ArrayList<Node> childNode;
	public Node parentNode;

	Node() {
		childNode = new ArrayList<Node>();
	}

}

  	 	 		 			 	 		 		    				 		