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.*;
import java.util.*;

public class Party {
    Scanner in = new Scanner(System.in);
    PrintStream out = System.out;
    HashMap<Integer, HashSet<Integer>> graph;
    
    void add(int source, int vertex) {
        HashSet<Integer> list = graph.get(source);
        list.add(vertex);
        graph.put(source, list);
    }
    
    void add(int v) {
        graph.put(v, new HashSet<Integer>());
    }
    
    int maxDepth(int v) {
        if (graph.get(v).isEmpty()) {
            return 1;
        }
        int max = 0;
        for (int hijo : graph.get(v)) {
            max = Math.max(max, maxDepth(hijo));
        }
        return 1 + max;
    }
    
    void solve() {
        graph = new HashMap<Integer, HashSet<Integer>>();
        ArrayList<Integer> root = new ArrayList<Integer>();
        int n = in.nextInt();
        for (int i = 1; i <= n; i++) {
            add(i);
        }
        for (int i = 1; i <= n; i++) {
            int p = in.nextInt();
            if (p != -1) {
                add(p, i);
            } else {
                root.add(i);
            }
        }
        int max = 0;
        for (int i = 0; i < root.size(); i++) {
            max = Math.max(max, maxDepth(root.get(i)));
        }
        out.println(max);
    }
    
    public static void main(String[] args) {
        new Party().solve();
    }
}