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.*;

public class file{
    public static void main(String[] args){
        Scanner scn = new Scanner(System.in);
        int n = scn.nextInt();
        int[] arr = new int[n];
        ArrayList<ArrayList<Integer>> graph = new ArrayList<>();
        for (int i = 0; i < n; i++) {
			graph.add(new ArrayList<>());
		}
        for(int i=0; i<n; i++){
            arr[i] = scn.nextInt();
            if(arr[i] != -1){
                arr[i]--;
                int u = i;
    			int v = arr[i];
    			graph.get(v).add(u);
            }
        }
        System.out.println(bfs(arr, graph));
    }
    
    public static class Pair{
        int val;
        int level;
        Pair(int v, int l){
            this.val = v;
            this.level = l;
        }
    }
    
    public static int bfs(int[] arr, ArrayList<ArrayList<Integer>> graph){
        LinkedList<Pair> queue = new LinkedList<>();
        HashSet<Integer> visited = new HashSet<>();
        for(int i=0; i<arr.length; i++){
            if(arr[i] == -1){
                queue.addLast(new Pair(i, 1));
            }
        }
        int max = 0;
        while(queue.size() > 0){
            Pair rem = queue.removeFirst();
            if(visited.contains(rem.val)){
                continue;
            }
            visited.add(rem.val);
            max = Math.max(max, rem.level);
            for(int i=0; i<graph.get(rem.val).size(); i++){
                queue.addLast(new Pair(graph.get(rem.val).get(i), rem.level + 1));
            }
        }
        return max;
    }
    
}