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.IOException;
import java.util.Scanner;

public class ProblemC {

    public static int[] memo;
    public static int[] manager;
    public static int[][] children;
    
    public static int length(int idx, int n) {
        if (memo[idx] > 0) {
            return memo[idx];
        }
        
        int max = 1;
        for (int i = 0 ; i < n ; i++) {
            if (children[idx][i] > 0) {
                max = Math.max(max, length(i, n) + 1);
            }
        }
        memo[idx] = max;
        
        return max;
    }

    public static void main(String[] args) throws IOException {
        Scanner s = new Scanner(System.in);
        int n = Integer.valueOf(s.nextLine());
        memo = new int[n];
        manager = new int[n];
        children = new int[n][n];
        for (int k = 0 ; k < n ; k++) {
            manager[k] = Integer.valueOf(s.nextLine());
            if (manager[k] >= 1) {
                manager[k] -= 1;
                children[manager[k]][k] = 1;
            }
        }
        
        int max = 0;
        for (int k = 0 ; k < n ; k++) {
            if (manager[k] == -1) {
                max = Math.max(max, length(k, n));
            }
        }
        System.out.println(max);
    }
}