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.HashSet;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;

public class Solution {
	public static void main(String args[]) {
		Scanner scan = new Scanner(System.in);
		int t = scan.nextInt();
		int n = t;
		scan.nextLine();
		boolean visited[] = new boolean[t + 1];
		Set<Integer> adj[] = new HashSet[t + 1];
		int a[] = new int[t + 1];
		for (int i = 0; i < t + 1; i++) {
			adj[i] = new HashSet<>();

		}
		int indx = 1;
		while (t > 0) {
			int v = Integer.parseInt(scan.nextLine());
			if (v != -1) {
				adj[indx].add(v);
				adj[v].add(indx);
			}
			a[indx] = v;
			indx++;
			t--;
		}
		int maxLevel = 1;
		for (int i = 1; i < visited.length; i++) {
			if (a[i] == -1) {
				Stack<Pair> s = new Stack<Pair>();
				s.push(new Pair(i, 1));
				visited[i] = true;
				while (!s.isEmpty()) {
					Pair r = s.pop();

					visited[r.getNode()] = true;
					maxLevel = Math.max(maxLevel, r.getLevel());
					for (Integer j : adj[r.getNode()]) {
						if (!visited[j]) {
							s.push(new Pair(j, r.getLevel() + 1));
						}
					}
				}
			}
		}
		System.out.println(maxLevel);
	}

	static class Pair {
		int node;
		int level;

		public Pair(int node, int level) {
			super();
			this.node = node;
			this.level = level;
		}

		public int getNode() {
			return node;
		}

		public void setNode(int node) {
			this.node = node;
		}

		public int getLevel() {
			return level;
		}

		public void setLevel(int level) {
			this.level = level;
		}

	}
}
