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 
//package codeforces;

import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.StreamTokenizer;
import java.util.ArrayList;

public class CodeForces {

    ArrayList<ArrayList<Integer>> tree = new ArrayList<ArrayList<Integer>>();
    int[] num;

    private int depth(int worker) {
        if (num[worker] == -1) {
            int max = 0;
            for (int i = 0; i < tree.get(worker).size(); i++) {
                max = Math.max(max, depth(tree.get(worker).get(i)));
            }
            num[worker] = max + 1;
        }
        return (num[worker]);
    }

    public void run() throws IOException {
        MyScanner sc = new MyScanner(System.in);
        int n = sc.nextInt();
        for (int i = 0; i < n; i++) {
            ArrayList<Integer> temp = new ArrayList<Integer>();
            tree.add(temp);
        }
        for (int i = 0; i < n; i++) {
            int boss = sc.nextInt() - 1;
            if (boss != -2) {
                ArrayList<Integer> temp = tree.get(boss);
                temp.add(i);
                tree.set(boss, temp);
            }
        }
        num = new int[n];
        for (int i = 0; i < n; i++) {
            num[i] = -1;
        }
        for (int i = 0; i < n; i++) {
            depth(i);
        }
        int max = 0;
        for (int i = 0; i < n; i++) {
            max = Math.max(max, num[i]);
        }
        System.out.println(max);
    }

    public static void main(String[] args) throws FileNotFoundException, IOException {
        CodeForces cf = new CodeForces();
        cf.run();
    }
}

class MyScanner {

    private StreamTokenizer st;

    public MyScanner(InputStream is) {
        st = new StreamTokenizer(is);
    }

    public MyScanner(File f) throws FileNotFoundException {
        BufferedReader in = new BufferedReader(new FileReader(f));
        st = new StreamTokenizer(in);
    }

    public int nextInt() throws IOException {
        st.nextToken();
        return ((int) st.nval);
    }

    public double nextDouble() throws IOException {
        st.nextToken();
        return (st.nval);
    }

    public String nextString() throws IOException {
        st.nextToken();
        return (st.sval);
    }
}
