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 Main {
    public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        new Main().run(in, out);
        out.close();
    }

    int N;
    List<Integer>[] adj;
    int[] depth;
    void run(FastScanner in, PrintWriter out) {

        N = in.nextInt();
        adj = new List[N];
        depth = new int[N];
        for (int i = 0; i < N; i++) adj[i] = new ArrayList<>();

        LinkedList<Integer> roots = new LinkedList<>();

        for (int v = 0; v < N; v++) {
            int u = in.nextInt()-1;
            if (u == -2) roots.add(v);
            else adj[u].add(v);
        }

        for (int root : roots) {
            dfs(root, 0);
        }

        out.println(maxDepth+1);
    }
    int maxDepth = 0;

    void dfs(int u, int d) {
        depth[u] = d;
        maxDepth = Math.max(maxDepth, depth[u]);
        for (int v : adj[u]) {
            dfs(v, d+1);
        }
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in));
            st = null;
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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