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.HashSet;
import java.util.Scanner;

public class Party {
	static int Max = 0;

	public static class Node {
		int index;
		ArrayList<Node> Neighbours = new ArrayList<>();
		boolean Visited;

		public Node(int index) {
			this.index = index;
		}
	}

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int N = scan.nextInt();
		Node[] nodes = new Node[N];
		HashSet<Node> Roots = new HashSet<Node>();
		for (int i = 0; i < N; i++)
			nodes[i] = new Node(i);
		for (int i = 0; i < N; i++) {
			int temp = scan.nextInt();
			if (temp == -1)
				Roots.add(nodes[i]);
			else {
				// nodes[i].Neighbours.add(nodes[temp - 1]);
				nodes[temp - 1].Neighbours.add(nodes[i]);
			}
		}
		DFS_LOOP(nodes, Roots);
		System.out.println(Max);
		scan.close();

	}

	private static void DFS_LOOP(Node[] nodes, HashSet<Node> roots) {
		// TODO Auto-generated method stub
		for (Node N : roots) {
			if (!N.Visited)
				DFS(N, 1);
		}
	}

	private static void DFS(Node n, int i) {
		// TODO Auto-generated method stub
		Max = Math.max(Max, i);
		n.Visited = true;
		for (Node nodeye : n.Neighbours) {
			if (!nodeye.Visited)
				DFS(nodeye, i + 1);
		}
	}
}
