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.InputStreamReader;
import java.io.IOException;
import java.util.ArrayList;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskC {

    ArrayList <Integer>[] g;
    
    int ans = 1;

    public void dfs (int v, int l){
        ans = Math.max(ans,l);
        for (int i=0; i<g[v].size(); i++){
            int to = g[v].get(i);
            dfs(to,l+1);
        }
    }

    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int n = Integer.parseInt(in.next());
        g = new ArrayList[n];
        for (int i=0; i<n; i++)
            g[i] = new ArrayList<Integer>();
        boolean[] boss = new boolean[n];
        for (int i=0; i<n; i++){
            int tval = Integer.parseInt(in.next());
            if (tval != -1){
                tval--;
                g[tval].add(i);
            }else {
                boss[i] = true;
            }
        }
        for (int i=0; i<n; i++)
            if (boss[i])
                dfs(i,1);
        out.print(ans);
    }
}

class InputReader {
    private BufferedReader reader;
    private StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream));
        tokenizer = null;
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException exception) {
                throw new RuntimeException(exception);
            }
        }
        return tokenizer.nextToken();
    }
}
