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.LinkedList;
import java.util.Scanner;

public class Party {

    static Boolean[] visited;
    static LinkedList<Integer>[] employee;
    static int maxDegree = 0, counter = 0, ans = 0;

    private static void DFS (int node){
        visited[node] = true;
        for (int i = 0; i < employee[node].size(); i++) {
            if (!visited[employee[node].get(i)]){
                counter++;
                DFS(employee[node].get(i));
            }
        }
        if (visited[node]){
            if (counter > maxDegree){
                maxDegree = counter;
            }
            counter--;
        }
    }

    public static void main(String[] args) throws IOException {
        Scanner br = new Scanner(System.in);

        int n = br.nextInt(), master;
        visited = new Boolean[n];
        employee = new LinkedList[n];
        LinkedList<Integer> roots = new LinkedList<>();

        for (int i = 0; i < n; i++) {
            visited[i] = new Boolean(false);
            employee[i] = new LinkedList<Integer>();
        }

        for (int i = 0; i < n; i++) {
            master = br.nextInt() - 1;
            if (master != -2){
                employee[master].add(i);
            } else {
                roots.add(i);
            }
        }

        for (int i = 0; i < roots.size(); i++) {
            DFS(roots.get(i));
            if (maxDegree > ans){
                ans = maxDegree;
            }
            counter = maxDegree = 0;
        }
        System.out.print(ans + 1);
    }
}
