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.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.ArrayList;
import java.util.PriorityQueue;
import java.util.Stack;
import java.util.TreeSet;


// this the solution for the third practical home work of DA.
// Q1 - Utopia company!
public class prog {
    // instead of scanner. for reading and printing data in console.
    StreamTokenizer st;
    PrintWriter pw;
        
    private ArrayList<Integer[]> result;
    private ArrayList<Integer>[] graph;
    private int n;
    private boolean[] isRoot;
    private PriorityQueue<Integer> Q;
    private int[] visited;
    private int[] depth;
    private int max;
    public prog(){
        st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        pw = new PrintWriter(new OutputStreamWriter(System.out));
        // prevents StreamTokenizer from reading next line tokens.
        st.eolIsSignificant(false);
        
        Q = new PriorityQueue<>();
    }
    // Read a token from console and cast it to a number
    private int nextInt() throws IOException
    {
        st.nextToken();
        return (int)st.nval;
    }
    
    public void Solv(){
        for (int i = 1; i<n+1; i++){
            if (isRoot[i] == true)
                BFS(i);
        }
        max = depth[1];
        for (int i = 1; i<n+1; i++)
            if(depth[i]>max)
                max = depth[i];
    }
    public void BFS(int j){
        //int depth;
        Q.clear();
        //Arrays.fill(visited, 0);
        visited[j] = 1;
        Q.add(j);
        while (!Q.isEmpty()){
            int u = Q.poll();
            for (int i = 0; i < graph[u].size(); i++)
                if (visited[graph[u].get(i)] == 0){
                    visited[graph[u].get(i)] = 1;
                    depth[graph[u].get(i)] = depth[u] + 1;
                    Q.add(graph[u].get(i));
                }
        }
        //return depth;
    }
    public void input(){
        try {
                n = nextInt();
                visited = new int[n+1];
                depth = new int[n+1];
                isRoot = new boolean[n+1];
                graph = new ArrayList[n+1];
                for (int i = 1; i<n + 1; i++ )
                    graph[i] = new ArrayList<Integer>();
                for (int i = 1; i < n +1; i++) {
                    int m = nextInt();
                    if ( m == -1){
                        isRoot[i] = true;
                    }
                    else
                        graph[m].add(i);
                }
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
    }
    public void output(){
        pw.println(max + 1);
        pw.flush();
    }
    public static void main(String[] args) {
        prog utopia = new prog();
        utopia.input();
        utopia.Solv(); // 1, 2 are always in sequence
        utopia.output();
    }
}