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.Scanner;

public class C87 {
    
    int[] emp;
    boolean[][] imm;
    int n;
    
    public void go() { 
        Scanner in = new Scanner(System.in);
        n =  Integer.parseInt(in.nextLine());
        imm = new boolean[n][n];
        emp = new int[n];
        
        for (int i = 0; i < n; i++) {
            int m =  Integer.parseInt(in.nextLine());
            emp[i] = m;
            if( m == -1) continue;
            imm[m-1][i] = true;     
        }
        
        int res = 0;
        
        for (int i = 0; i < n; i++) {
            if(emp[i] != -1) continue;
            
            int x = doit(i);
            
            if(res < x ) res = x;           
        }
        
        System.out.println(res);        
    
    
    
    
    }
    
    public int doit(int cur) {
        
        int res = 1;
        
        for (int i = 0; i < n; i++) {
            if( cur == i) continue;
            if(imm[cur][i]){
                int x = doit(i)+1;
                if(x > res) res = x;
            }
        }
        
        return res;
        
    }

    public static void main(String[] args) {
        new C87().go();     
    }

}
