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.Scanner;

/**
 * Created by akshaysharma on 21/11/16.
 */
public class Problem115A {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int[] source_arr = new int[N];
        ArrayList<ArrayList<Integer>> graph = new ArrayList<>();
        for (int i = 0; i < N; i++) {
            ArrayList<Integer> arr = new ArrayList<>();
            graph.add(i, arr);
        }
        boolean has_found = false;
        for (int i = 0; i < N; i++) {
            int index = sc.nextInt();
            if (index == -1) {
                source_arr[i] = -1;
            } else {
                ArrayList<Integer> arr = graph.get(index - 1);
                arr.add(i);
            }
        }
        boolean[] visited = new boolean[N];
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < N; i++) {
            if (!visited[i] && source_arr[i] == -1) {
                int res = findLevel(graph, visited, i);
                if (res > max)
                    max = res;
            }
        }
        System.out.print(max);
    }

    public static int findLevel(ArrayList<ArrayList<Integer>> graph, boolean[] visited, int root) {
        visited[root] = true;
        ArrayList<Integer> arr = graph.get(root);
        if (arr.size() == 0)
            return 1;

        int max = Integer.MIN_VALUE;
        int res = max;
        for (int node : arr) {
            res = findLevel(graph, visited, node);
            if (res > max) {
                max = res;
            }
        }
        return 1 + max;
    }

}
