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.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class ManagarialPart {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] indegree = new int[n + 1];
        LinkedList<Integer>[] G = new LinkedList[n + 1];
        for (int i = 1; i <= n; i++) {
            int m = in.nextInt();
            if (m != -1) {
                indegree[i] = 1;
                if (G[m] == null)
                    G[m] = new LinkedList<Integer>();
                G[m].add(i);
            }
        }
        int party = 0;
        int ppl = 0;
        Queue<Integer> current = new LinkedList<Integer>();
        Queue<Integer> next;
        for (int i = 1; i <= n; i++)
            if (indegree[i] == 0)
                current.add(i);
        while (ppl < n) {
            party++;
            next = new LinkedList<Integer>();
            while (!current.isEmpty()) {
                ppl++;
                int m = current.remove();
                if (G[m] != null) {
                    for (int ni : G[m]) {
                        next.add(ni);
                    }
                }
            }
            current = next;
        }
        System.out.println(party);

    }
}
