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.*;
import java.util.*;
public class Test {
  
    public static void main(String[] args) throws IOException{
      BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
      int n=Integer.parseInt(br.readLine());
      int[] p=new int[n+1];
      for(int i=1;i<=n;i++)
      {
        int x=Integer.parseInt(br.readLine());
        p[i]=x;
      }
      int[] l=new int[n+1];
      for(int i=1;i<=n;i++)
      {
        if(p[i]==-1)
         l[i]=0;
        else
        { int j=i;
          while(p[j]!=-1)
          {
            l[i]++;
            j=p[j];
          }
        } 
      }
      Arrays.sort(l);int count=0;
      for(int i=1;i<=n;i++)
      {
        int j=i;
        while(j<=n&&l[i]==l[j])
          j++;
        i=j-1;
        count++;  
      }
      System.out.println(count);
    }
}