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.*;

public class party {
    static boolean[] vis;
    static ArrayList<Integer>[] adjList;
    static int[] distance;
    static int employees;
    static int max;
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        employees = sc.nextInt();
        adjList = new ArrayList[employees];
        ArrayList<Integer> roots = new ArrayList<>();
        for(int i = 0; i<employees ; i++){adjList[i] = new ArrayList<>();}
        for(int i = 0; i<employees; i++){int x = sc.nextInt()-1; if(x!=-2){adjList[x].add(i);} else{roots.add(i);}}
        for(int root: roots){max = Math.max(max,bfs(root));}
        out.println(max+1);
        out.flush();

    }
    static int bfs(int start) {
        Queue<Integer> q = new LinkedList<>();
        q.add(start);
        int max = 0;
        distance = new int[employees];
        Arrays.fill(distance, -1);
        distance[start] = 0;
        while (!q.isEmpty()) {
            int cur = q.remove();
            for (int v : adjList[cur])
                if (distance[v] == -1) {
                    q.add(v);
                    distance[v] = distance[cur] + 1;
                    max = Math.max(max, distance[v]);
            }
        }
        return max;
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream system) {
            br = new BufferedReader(new InputStreamReader(system));
        }

        public Scanner(String file) throws Exception {
            br = new BufferedReader(new FileReader(file));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

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

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        public char nextChar() throws IOException {
            return next().charAt(0);
        }

        public Long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public boolean ready() throws IOException {
            return br.ready();
        }

        public void waitForInput() throws InterruptedException {
            Thread.sleep(3000);
        }
    }
}