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 Party {

	public static class Node {
		int id;
		Node parent;
		ArrayList<Node> children;
		
		public Node(int id) {
			this.id = id;
			children = new ArrayList<Node>();
		}
		
		public void addChild(Node node) {
			children.add(node);
		}
	}
	public static void main(String[] args) {
		Scanner stdIn = new Scanner(System.in);
		int N = stdIn.nextInt();
		int parent = -1;
		int max = Integer.MIN_VALUE;
		ArrayList<Node> roots = new ArrayList<Node>();
		
		Node []nodes = new Node[N];
		
		for (int i = 0; i < N; i++)
			nodes[i] = new Node(i + 1);
		
		for (int i = 0; i < N; i++) {
			parent = stdIn.nextInt();
			if (parent == -1) {
				roots.add(nodes[i]);
			} else {	
				nodes[i].parent = nodes[parent - 1];
				nodes[parent - 1].addChild(nodes[i]);
			}
		}
		
		for (Node root : roots) {
			max = Math.max(max, height(root));
		}
		System.out.println(max + 1);
		stdIn.close();
	}
	private static int height(Node root) {
		if (root == null)
			return -1;
		int max = -1;
		for (Node neighbor : root.children) {
			max = Math.max(height(neighbor), max);
		}
		return max + 1;
	}

}
