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.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class party {
	static final int VISITED = 2, EXPLORED = 1, UNVISITED = 0;
	static ArrayList<Integer> []adjList;
	static boolean[] vis ;
	static boolean group;
	public static void dfs(int u){
		vis[u]=true;
		for(int v: adjList[u])
			if(!vis[v])
				dfs(v);
		
		
	}
	
	static int[] parent, status;
	static void edgeType(int u)
	{
		status[u] = EXPLORED;
		for(int v: adjList[u])
			if(status[v] == UNVISITED)
			{
				System.out.printf("Edge from %d to %d is %s\n", u, v, "tree edge");
				parent[v] = u;
				dfs(v);
			}
			else
				if(status[v] == VISITED)	//cross edges only occur in directed graph
					group=false;
					else
					{
						System.out.printf("Edge from %d to %d is %s\n", u, v, "back edge");
						System.out.println("Cycle!");
					}
		status[u] = VISITED;
	}
	
	static int[] color;
	static boolean bipartitieCheck(int u)
	{
		for(int v: adjList[u])
			if(color[v] == -1)
			{
				color[v] = 1 ^ color[u];
				if(!bipartitieCheck(v))
					return false;
			}
			else
				if(color[v] == color[u])
					return false;
		return true;
	}
	
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		int n  = sc.nextInt();
		int[] x = new int[n];
		adjList = new ArrayList[n];
		for (int i = 0; i < x.length; i++) {
			adjList[i]=new ArrayList<>();
		}
		for (int i = 0; i < x.length; i++) {
			
		}
		for (int i = 0; i < x.length; i++) {
			int u = sc.nextInt();
			if(u!=-1){
				u--;
			int v =i;
			adjList[u].add(v);
			}
		}
		bfs_num=new int[n];
		for (int i = 0; i < n; i++) {
			vis=new boolean[n];
				bfs(i);
		}
		int mx=bfs_num[0];
		for (int i = 0; i < bfs_num.length; i++) {
			mx=Math.max(bfs_num[i], mx);
		}
	//	System.out.println(Arrays.toString(bfs_num));
		
		//System.out.println(mx+1);
		System.out.println(res+1);
		//int res=1;
		
		//System.out.println(Arrays.toString(color));
		
		
		
	}
	static int res;
	static int[] bfs_num;
	static void bfs(int s)
	{
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(s);
		vis[s] = true;
	 
		bfs_num[s]=0;
		while(!q.isEmpty())
		{
			int u = q.remove();
			for(int v: adjList[u])
				if(!vis[v])
				{
					vis[v] = true;
					q.add(v);
					bfs_num[v]=bfs_num[u]+1;
					res=Math.max(res, bfs_num[v]);
					//System.out.println("a7a");
				}
		}
	}
	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {return Integer.parseInt(next());}
		
		public long nextLong() throws IOException {return Long.parseLong(next());}

		public String nextLine() throws IOException {return br.readLine();}
		
		public double nextDouble() throws IOException
		{
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if(x.charAt(0) == '-')
			{
				neg = true;
				start++;
			}
			for(int i = start; i < x.length(); i++)
				if(x.charAt(i) == '.')
				{
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				}
				else
				{
					sb.append(x.charAt(i));
					if(dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg?-1:1);
		}
		
		public boolean ready() throws IOException {return br.ready();}


	}
}
