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.BufferedReader;
import java.io.InputStreamReader;
import java.util.Iterator;
import java.util.StringTokenizer;


public class Main {

static class node {

    public ArrayList<Integer> subnode = new ArrayList();
    public int layer;

}
    public static Scanner scn = new Scanner();
    public static ArrayList<Integer> root = new ArrayList<>();
    public static node arr[];

    public static void main(String[] args) {
        while (scn.hasNext()) {
            int node = scn.nextInt();
            arr = new node[node];
            for (int i = 0; i < node; i++) {
                arr[i] = new node();
            }
            for (int i = 0; i < node; i++) {
                int p = scn.nextInt();
                if (p != -1) {
                    arr[p - 1].subnode.add(i);
                } else {
                    root.add(i);
                    arr[i].layer = 1;
                }

            }
            int ans = 0;
            for (int i = 0; i < root.size(); i++) {
                int bfs = bfs(root.get(i));
                if (bfs > ans) {
                    ans = bfs;
                }
            }
            System.out.println(ans);
        }
    }

    public static int bfs(int i) {
        ArrayList<node> temp = new ArrayList();
        int ans = 1;
        temp.add(arr[i]);
        while (!temp.isEmpty()) {
            node cur = temp.remove(0);

            for (int j = 0; j < cur.subnode.size(); j++) {
                temp.add(arr[cur.subnode.get(j)]);
                arr[cur.subnode.get(j)].layer = cur.layer + 1;
                if(    arr[cur.subnode.get(j)].layer>ans)ans=    arr[cur.subnode.get(j)].layer;
            }

        }
        return ans;
    }

}

class Scanner implements Iterator<String> {

    BufferedReader buffer;
    StringTokenizer tok;

    public Scanner() {

        buffer = new BufferedReader(new InputStreamReader(System.in));

    }

    public boolean hasNext() {
        while (tok == null || !tok.hasMoreElements()) {
            try {
                tok = new StringTokenizer(buffer.readLine());
            } catch (Exception e) {
                return false;
            }
        }
        return true;
    }

    public String next() {
        if (hasNext()) {
            return tok.nextToken();
        }
        return null;
    }

    public void remove() {
        return;
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    long nextLong() {
        return Long.parseLong(next());
    }
}
