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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;

public class C {

	static int N;
	static ArrayList<Integer>[] Tree;
	static final int INF = (int)1e9;
	
	public static void main(String[] args) throws Exception{

		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		N = sc.nextInt();
		Tree = new ArrayList[N]; for(int i = 0; i < N; i++) Tree[i] = new ArrayList<>();
		boolean[] p = new boolean[N];
		for(int i = 0; i < N; i++)
		{
			int x = sc.nextInt() - 1;
			if(x == -2) p[i] = true;
			else
				Tree[x].add(i);
		}
		int ans = 0;
		for(int i = 0; i < N; i++)
			if(p[i])
				ans = Math.max(ans, BFS(i));
		out.println(ans + 1);
		out.flush();
		out.close();
	}
	
	static int BFS(int u)
	{
		int[] dist = new int[N];
		Arrays.fill(dist, INF);
		dist[u] = 0;
		Queue<Integer> q = new LinkedList<>();
		q.add(u);
		int max = 0;
		while(!q.isEmpty())
		{
			int cur = q.remove();
			for(int v : Tree[cur])
				if(dist[v] == INF)
				{
					q.add(v);
					dist[v] = dist[cur] + 1;
					max = Math.max(max, dist[v]);
				}
		}
		return max;
	}

	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream System){br = new BufferedReader(new InputStreamReader(System));}
		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public String nextLine()throws IOException{return br.readLine();}
		public int nextInt() throws IOException {return Integer.parseInt(next());}
		public double nextDouble() throws IOException {return Double.parseDouble(next());}
		public char nextChar()throws IOException{return next().charAt(0);}
		public Long nextLong()throws IOException{return Long.parseLong(next());}
		public boolean ready() throws IOException{return br.ready();}
		public void waitForInput(){for(long i = 0; i < 3e9; i++);}
	}
}