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.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class sample {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		boolean[][] adj = new boolean[n + 1][n + 1];
		int[] a = new int[n+1];
		for (int i = 0; i < n; i++) {
			a[i] = in.nextInt();
			if (a[i] != -1)
				adj[a[i]][i] = true;
		}
		int cnt = 0;
		Queue<Integer> cur = new LinkedList<Integer>();
		for (int i = 0; i <= n; i++)
			if (a[i] == -1)
				cur.add(i+1);
		int s;
		int j;
		while (!cur.isEmpty()) {
			cnt++;
			s = cur.size();
			for (int i = 0; i < s; i++) {
				j = cur.poll();
				for (int k = 0; k <= n; k++) {
					if (adj[j][k])
						cur.add(k+1);
				}
			}
		}
		System.out.println(cnt);
	}
}