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.*;

public class Problem116c{
	static class Node{
		int x, depth;
		public Node(int x, int depth){
			this.x = x;
			this.depth = depth;
		}
	}
	static HashMap<Integer, Integer> seniority;
	static HashMap<Integer, ArrayList<Integer>> children; 

	public static int dfs(int x, int deep){
		Stack<Node> s = new Stack<Node>();
		ArrayList<Integer> depths = new ArrayList<Integer>();
		s.push(new Node(x, deep+1));
		while(s.empty() == false){
			Node e = s.pop();
			if(children.get(e.x) == null){
				depths.add(e.depth);
			}
			else{
				//System.out.println("Putting in Children for: " + e.x);
				for(int c : children.get(e.x)){
					s.push(new Node(c, e.depth + 1));
					//System.out.println("Put in " + c+ " at depth: " + (e.depth+1));
				}

				//System.out.println("put in children for: "+x + " Current Depth: " + deep);
			}
			
		}

		int m = 0;
		for(int d : depths){
			if(d > m)
				m = d;
		}
		return m;
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int numEmps = sc.nextInt();
		seniority = new HashMap<Integer, Integer>();
		children = new HashMap<Integer, ArrayList<Integer>>();
		for(int i = 1; i <= numEmps; i++){
			int boss = sc.nextInt();
			seniority.put(i, boss);
			if(boss != -1){
				if(children.get(boss) == null){
					children.put(boss, new ArrayList<Integer>());
				}
				children.get(boss).add(i);
				
			}
			
		}

		ArrayList<Integer> depth = new ArrayList<Integer>();
		for(int a : seniority.keySet()){
			depth.add(dfs(a, 0));
		}
		int max = 0;
		for(int p : depth){
			//ystem.out.println(p);
			if(p > max)
				max = p;
		}
		System.out.println(max);
		
		/*System.out.println(seniority);
		for(int i : children.keySet()){
			System.out.print(i + ":  ");
			for(int k : children.get(i))
				System.out.print(k + "  ");
			System.out.println("");
		}*/
		
	}
}