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

//Problem 150A Party 
public class Main {

	static ArrayList<Integer> graph[];
	static int dist[];
	static boolean visited[];
	static int distMax=0;
	static boolean father[];
	
	public static void main(String[] args) {
		
		Scanner s = new Scanner(System.in);
		
		int n = s.nextInt();
		
		graph = new ArrayList[n];
		
		dist = new int[n];
		visited = new boolean[n];	
		father = new boolean[n];
		
		for(int i=0; i<n; i++) {
			graph[i] = new ArrayList<>();
			dist[i] = -1;
		}
				
		for(int i=0; i<n; i++) {
			int x = s.nextInt();	//0		
			
			if(x!=-1) { 
				x--;			
				graph[x].add(i);				
				father[i] = true;
			} 
		}	
		
		for(int i=0; i<n; i++) {			
			if(!father[i]) {
				bfs(i);
			}
		}		

		System.out.println(distMax+1);		
	}
	
	public static void bfs(int nodeInitial) {
		
		Queue<Integer> queue = new LinkedList<>();
		queue.add(nodeInitial);
		dist[nodeInitial] = 0;
		visited[nodeInitial] = true;
		
		while(!queue.isEmpty()) {
			
			int currentNode = queue.remove();
			
			for(int i=0; i<graph[currentNode].size(); i++) {
				
				int nodeNeighbour = graph[currentNode].get(i); 
				
				if(!visited[nodeNeighbour]) {
					
					visited[nodeNeighbour] = true;
					
					dist[nodeNeighbour] = dist[currentNode] + 1;
					
					if(dist[nodeNeighbour]>distMax) distMax = dist[nodeNeighbour];
					
					queue.add(nodeNeighbour);					
				}				
			}			
		}			
	}
	
}
