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

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

    static class CParty {
        int n;
        int[] adjList;
        boolean[] visited;

        public int dfs(int node) {
            if (node < 0)
                return 0;
            if (visited[node])
                return 0;
            visited[node] = true;
            return 1 + dfs(adjList[node]);
        }

        public void solve(int testNumber, Scanner sc, PrintWriter pw) {
            n = sc.nextInt();
            adjList = new int[n];
            for (int i = 0; i < n; i++)
                adjList[i] = sc.nextInt() - 1;
            int groups = 0;
            visited = new boolean[n];
            for (int i = 0; i < n; i++) {
                visited = new boolean[n];
                groups = Math.max(groups, dfs(adjList[i]));
            }
            pw.println(groups + 1);

        }

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            try {

                br = new BufferedReader(new InputStreamReader(s));
            } catch (Exception e) {
                throw new RuntimeException();
            }
        }

        public String next() {
            try {

                while (st == null || !st.hasMoreTokens())
                    st = new StringTokenizer(br.readLine());
                return st.nextToken();
            } catch (Exception e) {
                throw new RuntimeException();
            }
        }

        public int nextInt() {
            try {

                return Integer.parseInt(next());
            } catch (Exception e) {
                throw new RuntimeException();
            }
        }

    }
}

