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.PrintWriter;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.*;

public class C116 implements Runnable{
        
    public static void main(String args[]){
       new C116().run();
    }
  
    @Override
    public void run(){
        try{
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
            tok = null;
            solve();
            in.close();
            out.close();
            
        }
        catch(IOException e){
            e.printStackTrace();
            System.exit(0);
        }
    }
    
    int nextInt()throws IOException{
        return Integer.parseInt(nextToken());
    } 
    
    double nextDouble()throws IOException{
        return Double.parseDouble(nextToken());
    }
    
    long nextLong() throws IOException{
        return Long.parseLong(nextToken());
    }
    
    String nextToken()throws IOException{
        while(tok == null || !tok.hasMoreTokens()){
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }
    
    BufferedReader in;
    PrintWriter out;
    StringTokenizer tok;
    //////////////////////////////////////////////////////////////////
    //////////////////////////////////////////////////////////////////
    private void solve()throws IOException{
        int N = nextInt();
        int[] p = new int[N];
        int[] depth = new int[N];
        for(int i = 0; i < N; i++)p[i] = nextInt() - 1;
        
        int ans = 0;
        for(int i = 0; i < N; i++){
            depth[i] = calcdepth(i, p, depth);
            ans = Math.max(ans, depth[i]);
        }
        
        out.println(ans);
    }
    
    private int calcdepth(int at, int[] p, int[] depth){
        if(p[at] < 0) return 1;
        return calcdepth(p[at], p, depth) + 1;
    }
}

