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

public class Party {
    
    static ArrayList<Integer> [] adjList;
    static int V;
    static boolean [] vis;
    static int dist [];
    static int bfs(int s)
    {
        int last  = s;
        dist[s] = 1;
        vis[s] = true;
        Queue<Integer> q = new LinkedList<Integer>();
        q.add(s);
        while(!q.isEmpty())
        {
            int u = q.remove();
            for (int i = 0; i < adjList[u].size(); i++) {
            int v = adjList[u].get(i);
            if(!vis[v]){
            vis[v] = true;
            dist[v] = 1 + dist[u];
            q.add(v);
            last = v;
            }
            }
        }
        return dist[last];
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        adjList = new ArrayList[n];
        dist  = new int[n];
        vis = new boolean [n];
        for (int i = 0; i < n; i++) {
            adjList[i] = new ArrayList<Integer>();
        }
        ArrayList<Integer> roots = new ArrayList<Integer>();
        for (int i = 0; i < n; i++) {
            int u = sc.nextInt();
            if(u != -1)
            {
                adjList[u-1].add(i);
                adjList[i].add(u-1);
            }
            else
            {
                roots.add(i);
            }
        }
        int count = -1;
        for (int i = 0; i < roots.size(); i++) {
            if(!vis[roots.get(i)])
            {
                count = Math.max(bfs(roots.get(i)),count);
                Arrays.fill(dist, 0);
            }
        }
        System.out.println(count == 0 ? 1 : count);
    }
}