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 com.company;

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class Main {

    static int n;
    static ArrayList<Integer>[] w;
    static ArrayList<Integer> s;

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        n = Integer.parseInt(st.nextToken());
        w = new ArrayList[n + 1];
        for (int i = 0; i < n + 1; i++) {
            w[i] = new ArrayList<>();
        }
        s = new ArrayList<>();
        for (int i = 1; i < n + 1; i++) {
            st = new StringTokenizer(br.readLine() + " ");
            int a = Integer.parseInt(st.nextToken());
            if (a == -1) {
                s.add(i);
            } else {
                w[a].add(i);
                w[i].add(a);
            }
        }

        Queue<Integer> q = new LinkedList<>();
        boolean[] visited = new boolean[n + 1];
        int[] d = new int[n + 1];
        for (Integer i : s) {
            visited[i] = true;
            q.add(i);
            d[i] = 1;
        }
        int p;
        while (!q.isEmpty()) {
            p = q.remove();
            for (Integer i : w[p]) {
                if (!visited[i]) {
                    visited[i] = true;
                    d[i] = d[p] + 1;
                    q.add(i);
                }
            }
        }
        System.out.println(Arrays.stream(d).max().getAsInt());
    }
}
