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.*;

public class Codeforces{
	
	static class MyScanner{
		BufferedReader br;
		StringTokenizer st;
		
		MyScanner(){
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		String nn(){
			while(st == null || !st.hasMoreElements()){
				try{
					st = new StringTokenizer(br.readLine());
				}catch(IOException e){
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		
		int ni(){
			return Integer.parseInt(nn());
		}
		
		long nl(){
			return Long.parseLong(nn());
		}
		
		double nd(){
			return Double.parseDouble(nn());
		}
	}
	
	private static PrintWriter out;
	
	public static void main(String[] args){
		
		MyScanner sc = new MyScanner();
		out = new PrintWriter(new BufferedOutputStream(System.out));
		
		int n = sc.ni();
		
		HashMap<Integer, Integer> hashMap = new HashMap<>();
		
		for(int i = 0; i < n; i++){
			int a = sc.ni();
			hashMap.put(i + 1, a);
		}
		
		int[] ar = new int[n + 1];
		
		// System.out.println(hashMap);
		
		for(int i = 1; i <= n; i++){
			if(ar[i] == 0)
				fillArray(hashMap, ar, n, i);
		}
		
		int ans = 0;
		
		for(int i = 1; i <= n; i++) ans = Math.max(ans, ar[i]);
		
		out.println(ans);
		
		// System.out.println(Arrays.toString(ar));
		
		out.close();
	}
	
	private static int fillArray(HashMap<Integer, Integer> hashMap, int[] ar, int n, int cur){
		if(hashMap.get(cur) == -1){
			ar[cur] = 1;
			return ar[cur];
		}
		
		if(ar[cur] > 0)
			return ar[cur];
		
		ar[cur] = 1 + fillArray(hashMap, ar, n, hashMap.get(cur));
		return ar[cur];
	}
}