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.IOException;
import java.io.InputStreamReader;
import java.util.LinkedList;
import java.util.Queue;

public class Party {
    public static void main(String[] args) throws NumberFormatException,
            IOException {
        BufferedReader buf = new BufferedReader(
                new InputStreamReader(System.in));
        int n = Integer.parseInt(buf.readLine());
        Queue<Integer> S = new LinkedList<Integer>();
        int[] commingEdges = new int[n];
        LinkedList<Integer>[] L = new LinkedList[n];
        for (int i = 0; i < n; i++)
            L[i] = new LinkedList<Integer>();
        for (int i = 0; i < n; i++) {
            int x = Integer.parseInt(buf.readLine());
            if (x != -1) {
            x-=1;
            
                commingEdges[i]++;
                L[x].add(i);
            }

        }
        for (int i = 0; i < n; i++)
            if (commingEdges[i] == 0) {
                S.add(i);
                S.add(1);

            }
        int count = 1;
        int level = 0;
        while (!S.isEmpty()) {

            int a = S.poll();
            level = S.poll();
            
            for (Integer i : L[a]) {
                commingEdges[i]--;
                if (commingEdges[i] == 0) {
                    S.add(i);
                    S.add(level + 1);
                }
            }
        }
        System.out.println(level);
    }

}
