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.Scanner;
import java.io.PrintWriter;

public class Main {
    
    static Scanner s = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);
    static int n = s.nextInt(), l[] = new int[n], m[] = new int[n];
    
    public static int Something(int i) {
        if(i == -1) return 0;
        if(l[i] != 0) return l[i];
        l[i] = Something(m[i]) + 1;
        return l[i];
    }
    
    public static void main(String[] args) {
        for(int i = 0; i < n; i++) {
            m[i] = Math.max(0, s.nextInt())-1;
        }
        int r = 0;
        for(int i = 0; i < n; i++) {
            if(l[i] == 0) Something(i);
            r = Math.max(r, l[i]);
        }
        out.print(r);
        out.close();
    }
}