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

public class Party {

    public static boolean conec[][];
    public static boolean seen[];
    public static int n = 0;
    public static int ans = 0;
    public static int num = 0;
    public static int color[];
    public static int parent[];
    public static int distance[];

    public void solve() throws IOException {

        Scanner sc = new Scanner(System.in);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        n = sc.nextInt();
        int i = 0, j = 0, k = 0;

        conec = new boolean[n][n];
        seen = new boolean[n];
        color = new int[n];
        distance = new int[n];
        parent = new int[n];

        int ga[] = new int[n];
        for (i = 0; i < n; ++i) {
            int tem = sc.nextInt();
            ga[i]=tem;
            if (tem != -1) {
                // conec[i][tem - 1] = true;
                conec[tem - 1][i] = true;
            }
        }

        int now = 0;
        for (j = 0; j < n; ++j) {
            if (ga[j]==-1) {
                bfs(j);
                //System.out.println(Arrays.toString(distance));
                for (i = 0; i < n; ++i) {
                    if (now < distance[i]) {
                        now = distance[i];
                    }
                }
                seen = new boolean[n];
                color = new int[n];
                distance = new int[n];
                parent = new int[n];
            }
        }
        System.out.println(now + 1);
    }

    public void bfs(int s) {

        Queue<Integer> q = new LinkedList<Integer>();
        color[s] = 1;
        distance[s] = 0;
        parent[s] = -1;
        q.add(s);

        while (!q.isEmpty()) {
            int u = q.poll();

            for (int v = 0; v < n; ++v) {
                if (color[v] == 0 && conec[u][v]) {
                    color[v] = 1;
                    distance[v] = distance[u] + 1;
                    parent[v] = u;
                    q.add(v);
                }
            }
            color[u] = 2;

        }

    }

    public static void main(String[] args) throws IOException {
        new Party().solve();
    }
}
