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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;


public class ProblemA {

	private static class Node {
		List<Integer> edges = new ArrayList<Integer>();
	}
	
	Node[] graph;
	
	public void solve() throws IOException {
		BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        
        int n = Integer.valueOf(stdin.readLine());
        graph = new Node[n];
        for (int i = 0; i < n; i++) {
        	graph[i] = new Node();
        }
        
        List<Integer> start = new ArrayList<Integer>();
        for (int i = 0; i < n; i++) {
        	int p = Integer.valueOf(stdin.readLine());
        	if (p == -1) {
        		start.add(i);
        	} else {
        		graph[p-1].edges.add(i);
        	}
        }
        
        int max = -1;
        for (int v : start) {
        	max = Math.max(max, path(v));
        }
        
        out.println(max);
        
        out.flush();
        out.close();
	}
	
	private int path(int v) {
		int max = 0;
		for (int u : graph[v].edges) {
			max = Math.max(max, path(u));
		}
		return max + 1;
	}

	public static void main(String[] args) throws IOException {
        ProblemA solver = new ProblemA();
        solver.solve();
	}
}
