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.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Scanner;


public class C_116C {
	private static ArrayList<Integer> edges[];
	private static ArrayList<Integer> TopSort;
	private static boolean[] used;
	private static int[] d;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		edges = new ArrayList[n+1];
		for (int i = 1; i <=n; i++) {
			edges[i] = new ArrayList<Integer>();
		}
		for (int i = 1; i <=n; i++) {
			int sur = sc.nextInt();
			if(sur!=-1) edges[sur].add(i);
		}
		TopSort = new ArrayList<Integer>();
		used = new boolean[n+1];
		int ans = 0;
		for (int i = 1; i <=n; i++) {
			if(!used[i]){
				dfs(i);
			}
		}
		Arrays.fill(used, false);
		d = new int[n+1];
		Collections.reverse(TopSort);
		for (int v : TopSort) {
			if(!used[v]) {
				d[v] = 1;
				dfs2(v);
			}
		}
//		System.out.println(TopSort);
		for (int i = 1; i <=n; i++) {
//			System.out.print(d[i]+" ");
			ans = Math.max(ans, d[i]);
		}
		System.out.println(ans);
	}
	private static void dfs2(int v) {
		used[v] = true;
		for (int i : edges[v]) {
			if(!used[i]){
				d[i] = d[v] + 1;
				dfs2(i);
			}
		}
	}

	private static void dfs(int v) {
		used[v] = true;
		for (int i : edges[v]) {
			if(!used[i]){
				dfs(i);
			}
		}
		TopSort.add(v);
	}
}
