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.*;
import java.io.*;
import java.math.*;

public class Main {
    private static final Scanner scanner = 
        new Scanner(new BufferedReader(new InputStreamReader(System.in)));

    public static void main(String[] args) {
        Main solution = new Main();
        solution.run();
    }
    
    public void run() {
        int n = scanner.nextInt();

        int[] parent = new int[n];
        int[] dist = new int[n];

        Arrays.fill(dist, -1);

        for (int i = 0; i < n; i++) {
            parent[i] = scanner.nextInt();
            if (parent[i] == -1) {
                dist[i] = 0;
            } else {
                parent[i]--;
            }
        }

        int max = 0;
        for (int i = 0; i < n; i++) {
            dist[i] = dfs(i, parent, dist);
            max = Math.max(max, dist[i] + 1);
        }

        System.out.println(max);
    }

    private int dfs(int i, int[] parent, int[] dist) {
        if (dist[i] != -1) {
            return dist[i];
        } else {
            return dist[i] = dfs(parent[i], parent, dist) + 1;
        }
    }
}
