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.PrintWriter;
import java.util.*;

public class Main{
    public static BufferedReader in;
    public static int gi(String s) throws IOException {
        return Integer.parseInt(s);
    }
    static int n;
    static boolean avoid[][], used[];
    static int maxg=0;
    static void dfs(int x, int g){
        used[x]=true;
        maxg=Math.max(g,maxg);
        for(int i=0; i<n; i++){
            if(used[i]||x==i||!avoid[x][i]) continue;
            dfs(i, g+1);
        }
    }
    public static void main(String args[]) throws Exception{
        PrintWriter out = new PrintWriter(System.out);
        in = new BufferedReader(new InputStreamReader(System.in));
        n = Integer.parseInt(in.readLine());
        int p[] = new int[n];
        avoid = new boolean[2000][2000];
        int dfsin[] = new int[n];
        int w=0;
        for(int i=0; i<n; i++){
            p[i] = gi(in.readLine())-1;
            if(p[i]<0) 
            {
                dfsin[w++] = i;                 
                continue;
            }
            avoid[p[i]][i]=true;
        }
        boolean globalused[] = new boolean[n];
        for(int k=0; k<w;k++){
            int i = dfsin[k];
            if(globalused[i]) continue;
            used=new boolean[n];
            dfs(i,1);
            for(int j=0; j<n; j++){
                if(used[j])
                globalused[j]=true;
            }
        }
        out.print(maxg);
        out.close();
    }
}