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 static java.util.Arrays.*;
import static java.lang.Math.*;

// Codeforces Beta Round #87 Div2 C

public class Main {
    static Scanner sc = new Scanner(System.in);
    
    public static void main(String[] args) {
        int n = sc.nextInt();
        Tree[] ts = new Tree[n];
        for (int i = 0; i < n; i++) 
            ts[i] = new Tree(i + 1, sc.nextInt());
        
        int res = 0;
        for (int i = 0; i < n; i++) { 
            int r, cur = i + 1;
            for (r = 1; ts[cur - 1].par != -1; cur = ts[cur - 1].par, r++) {
                //System.out.println(cur);
            }
            res = max(res, r);
        }
        System.out.println(res);
    }
}

class Tree {
    int no, par;
    Tree(int no, int par) {
        this.no = no;
        this.par = par;
    }
}
