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.io.*; import java.math.*; import java.text.*;

public class main {
	
	static Scanner scan = new Scanner(System.in);

	public static void main(String[] args) throws IOException{
		int N = scan.nextInt(), answer = Integer.MIN_VALUE;
		graph g = new graph(); g.init(N); boolean visited[] = new boolean[N];
		ArrayList<Integer> query = new ArrayList<>(); 
		for(int i=0; i<N; i++) {
			int temp = scan.nextInt();
			if(temp == -1) {query.add(i);}
			else {g.addEdge(temp-1, i);}
		}
		for(int i=0; i<query.size(); i++) {
			LinkedList<Integer> queue = new LinkedList<>(); queue.add(query.get(i)); visited[query.get(i)] = true;
			//System.out.println(g.bfs(queue, visited));
		    answer = Integer.max(answer, g.bfs(queue, visited));
		}
		System.out.println(answer);
	}
	
	static class graph{
		public HashMap<Integer, HashMap<Integer, Integer>> map = new HashMap<>();
		public void init(int N) {
			for(int i=0; i<N; i++) {
				map.put(i, new HashMap<>());
			}
		}
		public void addEdge(int src, int des) {
			if(!map.containsKey(src)) {map.put(src, new HashMap<>());}
			if(!map.containsKey(des)) {map.put(des, new HashMap<>());}
			map.get(src).put(des, 1);
		}
		public int bfs(LinkedList<Integer> queue, boolean[] visited) {
			int temp = 1, answer = 0;
			while(!queue.isEmpty()) {
				//System.out.println(queue);
				int temp2 = 0; 
				for(int i=0; i<temp; i++) {
					int src = queue.removeFirst(); Set<Integer> set = map.get(src).keySet();
					for(int k : set) {
						if(!visited[k]) {
							visited[k] = true; queue.addLast(k); temp2++; 
						}
					}
				}
				answer++;
				temp = temp2;
			}
			
			return answer;
		}
	}
	
}
