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.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class test {
	static ArrayList<Integer> adjList[];
	static boolean visited[];
	static int level [];

	static void bfs(int start, int lev) {
		Queue<Integer> set = new LinkedList<>();
		set.add(start);
		level[start] = lev;
		visited[start] = true;
		while (!set.isEmpty()) {
			int u = set.remove();
			for (int v : adjList[u])
				if (!visited[v]) {
					visited[v] = true;
					set.add(v);
					level[v] = level[u] + 1;
				}
		}
	}

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		ArrayList<Integer>start = new ArrayList<>();
		adjList = new ArrayList[n];
		visited = new boolean[n];
		level = new int[n];
		Arrays.fill(visited, false);
		for (int i = 0; i < n; i++) {
			adjList[i] = new ArrayList<>();
		}
		for (int i = 0; i < n; i++) {
			int u = sc.nextInt();
			if (u != -1){
				adjList[u-1].add(i);
			    //adjList[i].add(u-1);
			}else
				start.add(i);
		}
		while(!start.isEmpty()){
        bfs(start.remove(0),1);
        
		}
        Arrays.sort(level);
        System.out.println(level[n-1]);
	}
}