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 
// https://codeforces.com/problemset/problem/115/A
// rohchakr
//

import java.util.Scanner;
import java.util.LinkedList;
import java.util.ArrayList;
import java.util.ArrayDeque;

public class Main {
  public static void main (String [] args) {
    Scanner s = new Scanner (System.in);
    int n = s.nextInt();
    int [] p = new int[n+1];
    for (int i = 1; i<=n; i++)
      p[i] = s.nextInt();
    
    // all p[i] with -1 can be considered as roots,
    // need to find the max length from any root to its leaves    
    ArrayList<LinkedList<Integer> > adjacencyList = new ArrayList<LinkedList<Integer> >(n+1);
    for (int i = 0; i<=n; i++) {
      adjacencyList.add(i,new LinkedList<Integer>());
    }
    for (int i = 1; i<=n; i++) {
      if(p[i] != -1)
        adjacencyList.get(p[i]).add(i);
    }
    //System.out.println(adjacencyList.toString());
   
    // bfs for all i whose p[i] = -1
    int maxReportsAll = 0;
    for (int i = 1; i <= n; i++) {
      if(p[i] == -1) {
        int maxReports = getMaxReports(i,adjacencyList);
        //System.out.println("MaxReports-" + i + "]   =    " + maxReports);
        maxReportsAll = (maxReportsAll > maxReports) ? maxReportsAll : maxReports;
      }
    }
    System.out.println(maxReportsAll);
  }

  private static int getMaxReports(int root, ArrayList<LinkedList<Integer> > adjacencyList) {
    ArrayDeque<Integer> reports = new ArrayDeque<Integer>();
    reports.add(root);
    int length = 0;
    while(reports.size()>0){
        int reportsSize = reports.size();
        for(int i = 0; i < reportsSize; i++) {
          int popped = reports.remove();
          for(int j = 0; j<adjacencyList.get(popped).size(); j++)
            reports.add(adjacencyList.get(popped).get(j));
      }
      length++;
    }
    return length;
  }
}
