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.*;
import java.lang.*;
import java.io.*;

public class Solution{
    
    public static void dfs(ArrayList<Integer> arr[],int node,int[] isVis,int h){
        isVis[node] = 1;
        h++;
        height = Math.max(height,h);
        //System.out.println(h+" "+"hello");
        for(int i=0;i<arr[node].size();i++){
            if(isVis[arr[node].get(i)] == 0)
                dfs(arr,arr[node].get(i),isVis,h);
        }
    }
    static int height = 0;
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        ArrayList<Integer> arr[] = new ArrayList[n+1];
        for(int i=1;i<=n;i++){
            arr[i] = new ArrayList<Integer>();
        }
        int arr2[] = new int[n+1];
        for(int i=1;i<=n;i++){
            int a = sc.nextInt();
            arr2[i] = a;
            if(a != -1)
            {   
                arr[a].add(i);
                arr[i].add(a);
            }
        }
        /*for(int i=1;i<=n;i++){
            
            for(int j=0;j<arr[i].size();j++){
                System.out.print(arr[i].get(j)+" ");
            }
            System.out.println("");
        }*/
        int isVis[] = new int[n+1];
        int ans = 0;
        for(int i=1;i<=n;i++){
            if(isVis[i] == 0 && arr2[i] == -1)
            {   height = 0;
                dfs(arr,i,isVis,0);
                ans = Math.max(ans,height);
            }
        }
        System.out.println(ans);    
    }     
}