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.BufferedReader;
import java.io.InputStreamReader;
import java.io.*;
import java.lang.Integer;
import java.util.*;

public class Party {
    public static int max = 0;
    public static void main (String[] args) throws Exception {
        InputStreamReader isr = new InputStreamReader(System.in);
        BufferedReader br = new BufferedReader(isr);
        int size = Integer.parseInt(br.readLine());
        int[] parents = new int[size];
        for (int i = 0; i < size; i++) {
            parents[i] = Integer.parseInt(br.readLine());
        }
        int num = 1;
//        for (int a : parents) {
//            System.out.println("a" + a);
//        }
        for (int index = 0; index < parents.length; index++) {
            recurse(parents, num, index);
        }
        System.out.println(max);
    }
    public static void recurse(int[] parents, int num, int index) {
        if (parents[index] == -1) {
            if (num > max) {
                max = num;
                //System.out.println(max);
            }
            return;
        }
        //System.out.println(num);
        num = num + 1;
        index = parents[index] - 1;
        //System.out.println(index);
        recurse(parents, num, index);
    }
}