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;
import java.util.List;
import java.util.StringTokenizer;


public class ProblemC implements Runnable {
	int mark[];
	ArrayList<Integer>[] graph;
	int count = 0;
	int count1 = 0;
	private void solve() throws IOException {
		int n = nextInt();
		 graph = new ArrayList[n];
		for (int i = 0; i < n; ++i) {
			graph[i] = new ArrayList<Integer>();
		}
		List<Integer> v = new ArrayList<Integer>();
		for (int i = 0; i < n; ++i) {
			int a = nextInt() - 1;
			if (a == -2) {
				v.add(i);
			} else {
				graph[i].add(a);
				graph[a].add(i);
			}
		}
		mark = new int[n];
		for (int x : v) {
			count1 = 0;
			dfs(x, 1);
			count = Math.max(count, count1);
		}
		writer.println(count);
	}
	
	private void dfs(int x, int depth) {
		mark[x] = 1;
		count = Math.max(count, depth);
		for (int i = 0; i < graph[x].size(); ++i) {
			int w = graph[x].get(i);
			if (mark[w] != 1) {
				dfs(w, depth + 1);
			}
		}
	}
	
	StringTokenizer tokenizer;
	BufferedReader reader;
	PrintWriter writer;
	@Override
	public void run() {
		try {
			tokenizer = null;
			reader = new BufferedReader(new InputStreamReader(System.in));
			writer = new PrintWriter(System.out);
			solve();
			reader.close();
			writer.close();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}
	
	private String nextToken() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			tokenizer = new StringTokenizer(reader.readLine());
		}
		return tokenizer.nextToken();
	}
	
	private int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}
	
	private long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}
	
	private double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
	
	public static void main(String args[]) {
		new ProblemC().run();
	}

}
