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.*;

public class C {
	
	static boolean[] used;
	static List<List<Integer>> g;
	public static void dfs(int v) {
		used[v] = true;
		for (int i = 0; i < g.get(v).size(); i++) {
			if (!used[g.get(v).get(i)]) {
				if (g.get(g.get(v).get(i)).size() != 0) {
					dfs(g.get(v).get(i));
				} else {
					g.get(v).remove(i);
					i--;
				}
				
			}
		}
	}
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		g = new ArrayList<List<Integer>>(n + 1);
		List<Integer> roots = new ArrayList<Integer>();
		used = new boolean[n + 1];
		for (int i = 0; i <= n; i++) g.add(new ArrayList<Integer>());
		for (int i = 0; i < n; i++) {
			int x = sc.nextInt() - 1;
			if (x == -2) {
				roots.add(i);
			} else {
				g.get(x).add(i);
			}
		}
		int k = 0;
		for (int i = 0; i < roots.size(); i++) {
			//if (g.get(roots.get(i)).size() != 0) {
				g.get(n).add(roots.get(i));
			//}
		}
		while (g.get(n).size() != 0) {
			used = new boolean[n + 1];
			dfs(n);
			++k;
		}
		System.out.print(k);
	}
}
