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.Scanner;

/**
 * Created by Baka on 30.12.2016.
 * http://codeforces.com/problemset/problem/115/A
 */
public class Holiday {
    static boolean visit[];
    static int max = 0;

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int v = scanner.nextInt();
        int p[] = new int[v];
        visit = new boolean[v];

        for (int i = 0; i < v; i++) {
            p[i] = scanner.nextInt();
        }

        ArrayList<Integer> g[] = new ArrayList[v];
        for (int i = 0; i < v; i++) {
            g[i] = new ArrayList<Integer>();
        }

        for (int i = 0; i < v; i++) {
            if (p[i] == -1) continue;
            g[p[i] - 1].add(i);
        }

        for (int i = 0; i < v; i++) {
            if (p[i] == -1) dfs(g, i, 1);
        }

        System.out.println(max);
    }

    static void dfs(ArrayList<Integer> graph[], int v, int depth) {
        max = Math.max(max, depth);
        visit[v] = true;
        for (int i : graph[v]) {
            if (!visit[i])
                dfs(graph, i, depth + 1);
        }
    }

}
