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 class DirectedGraph {
        private ArrayList<Integer> [] adjLists;
        private int [] level;
        
        DirectedGraph(int nVertex) {
            adjLists = (ArrayList<Integer> []) new ArrayList [nVertex];
            for(int i = 0; i < nVertex; i++)
                adjLists[i] = new ArrayList<Integer>();
            
            level = new int [nVertex];
            for(int i = 0; i < nVertex; i++)
                level[i] = -1;
        }
        
        void addEdge(int from, int to) {
            adjLists[from].add(to);
        }
        
        void setLevels() {
            for(int i = 0; i < level.length; i++) 
                level[i] = getLevel(i);
        }
        
        int getLevel(int v) {
            if(level[v] != -1) 
                return level[v];
            
            int maxLevel = 0;
            ArrayList<Integer> children = adjLists[v];
            for(Integer child : children) {
                int level = getLevel(child);
                if(level > maxLevel)
                    maxLevel = level;
            }
            
            return maxLevel + 1;        
        }
        
        int getMaxLevel() {
            int maxLevel = 0;
            for(int i : level) 
                if(i > maxLevel)
                    maxLevel = i;
            return maxLevel;
        }
    }
    
    
    public static void main(String [] args) {
        Scanner r = new Scanner(System.in);
        int nVertex = r.nextInt();
        DirectedGraph g = new DirectedGraph(nVertex);
        
        for(int i = 0; i < nVertex; i++) {
            int from = r.nextInt();
            if(from != -1) 
                g.addEdge(from - 1, i);
        }
        
        g.setLevels();
        System.out.println(g.getMaxLevel());
    }
}