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.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Map<Integer, List<Integer>> graph = new HashMap<>();
        List<Integer> roots = new ArrayList<>();
        Scanner sc = new Scanner(System.in);
        int emloyeeCount = sc.nextInt();
        for (int i = 0; i < emloyeeCount; i++) {
            int boss = sc.nextInt();
            if (boss == -1) {
                roots.add(i + 1);
            } else {
                if (!graph.containsKey(boss)) {
                    graph.put(boss, new ArrayList<>());
                }
                graph.get(boss).add(i + 1);
            }
        }
        int maxDepth = 1;
        for (Integer root: roots) {
            int depth = dfs(root, graph, 1);
            maxDepth = depth > maxDepth ? depth : maxDepth;
        }
        System.out.println(maxDepth);
    }

    private static int dfs(Integer boss, Map<Integer, List<Integer>> graph, int depth) {
        if (!graph.containsKey(boss)) {
            return depth;
        }
        int maxDepth = depth;
        for (Integer employee: graph.get(boss)) {
            int newDepth = dfs(employee, graph, depth + 1);
            maxDepth = newDepth > maxDepth ? newDepth : maxDepth;
        }
        return maxDepth;
    }
}