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.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;

public class Party {
	static ArrayList<Integer>[] adjlist;
	static boolean[] vis;
	static int[] hs;
	@SuppressWarnings("unchecked")
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		int v = Integer.parseInt(br.readLine());
		adjlist = new ArrayList[v+1];
		for(int i=0;i<adjlist.length;i++)
			adjlist[i] = new ArrayList<>();
		vis = new boolean[v+1];
		hs = new int[v+1];
		for(int i=1;i<=v;i++) {
			int par = Integer.parseInt(br.readLine());
			if(par!=-1)
				adjlist[par].add(i);
		}
		for(int i=1;i<vis.length;i++) {
			if(!vis[i]) {
				vis = new boolean[v+1];
				dfs(i,0);
			}
		}
		ArrayList<Integer> all = new ArrayList<>();
		for(int i=0;i<hs.length;i++) {
			if(!all.contains(hs[i]))
				all.add(hs[i]);
		}
		pw.println(all.size());
		pw.close();
	}
	
	public static void dfs(int v,int lvl) {
		vis[v] = true;
		hs[v] = Math.max(lvl, hs[v]);
		for(int i=0;i<adjlist[v].size();i++) {
			if(!vis[adjlist[v].get(i)]) {
				dfs(adjlist[v].get(i),lvl+1);
			}
		}
	}

}