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 Main {
    static ArrayList<Integer>[] adjList;
    static boolean[] visited;
    static int n;
    static int bfs(int s){
        Queue<Integer> q = new LinkedList<>();
        q.add(s);
        visited[s] = true;
        int[] dist = new int[n];
        while(!q.isEmpty()){
            int u = q.poll();
            for (int v : adjList[u]) {
                q.add(v);
                visited[v] = true;
                dist[v] = dist[u] + 1;
            }
        }

        int max = 0;
        for (int x : dist) {
            max = Math.max(max, x);
        }
        return max;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        adjList = new ArrayList[n];
        visited = new boolean[n];
        for (int i = 0; i < n; i++) {
            adjList[i] = new ArrayList<>();
        }

        for (int i = 0; i < n; i++) {
            int x = sc.nextInt();
            if(x != -1){
                adjList[x-1].add(i);
            }
        }
        int ans = 0;
        for (int i = 0; i < n; i++) {
            ans = Math.max(ans, bfs(i));
        }

        System.out.println(ans + 1);
    }
}
   			 		  	 			  	  	 	      	