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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.StringTokenizer;

public class A115 {

	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer st = null;

	int ni() throws Exception {
		if (st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(br.readLine());
		}
		return Integer.parseInt(st.nextToken());
	}

	int n, answer = Integer.MIN_VALUE;
	List<Integer>[] employees;
	boolean[] visited;
	Queue<Integer> notSupervised = new LinkedList<Integer>();

	A115() throws Exception {
		n = ni();
		employees = new List[n + 1];
		for (int i = 1; i <= n; i++) {
			employees[i] = new ArrayList<>();
		}
		visited = new boolean[n + 1];
		for (int employee = 1; employee <= n; employee++) {
			int supervisor = ni();
			if (supervisor != -1) {
				employees[supervisor].add(employee);
			} else {
				notSupervised.add(employee);
			}
		}
		while (!notSupervised.isEmpty()) {
			int cur = notSupervised.poll();
			if (!visited[cur]) {
				height(cur, 1);
			}
		}
		System.out.println(answer);
	}

	void height(int supervisor, int level) {
		if (!visited[supervisor]) {
			visited[supervisor] = true;
			if (employees[supervisor].isEmpty()) {
				answer = Math.max(answer, level);
			} else {
				for (int employee : employees[supervisor]) {
					if (!visited[employee]) {
						height(employee, level + 1);
					}
				}
			}
		}
	}

	public static void main(String[] args) throws Exception {
		new A115();
	}

}
