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 
/**
 * Created by ankeet on 7/17/16.
 */
import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class A115 {

    static FastReader in = null;
    static PrintWriter out = null;
    static int[] dep = null;
    static int[] p =null;
    public static void fill(int curr)
    {
        if(dep[curr] != 0) return;
        if(p[curr] == -1) {
            dep[curr] = 1;
            return;
        }
        if(dep[p[curr]] == 0)
        {
            fill(p[curr]);
        }
        dep[curr] = dep[p[curr]]+1;

    }
    public static void solve()
    {
        int n = in.nint();
        p = new int[n+1];
        dep = new int[n+1];
        for(int i=1; i<=n; i++) p[i] = in.nint();
        int max = 0;
        for(int i=1; i<=n; i++)
        {
            fill(i);
            max = Math.max(max, dep[i]);
        }
        out.println(max);
    }

    public static void main(String[] args)
    {
        in = new FastReader(System.in);
        out = new PrintWriter(System.out);
        solve();
        out.flush();
        out.close();

    }

    static class FastReader {

        BufferedReader read;
        StringTokenizer tokenizer;

        public FastReader(InputStream in)
        {
            read = new BufferedReader(new InputStreamReader(in));
        }

        public String next()
        {
            while(tokenizer == null || !tokenizer.hasMoreTokens())
            {
                try{
                    tokenizer = new StringTokenizer(read.readLine());
                }catch(Exception e){
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nint()
        {
            return Integer.parseInt(next());
        }
        public long nlong()
        {
            return Long.parseLong(next());
        }
        public double ndouble()
        {
            return Double.parseDouble(next());
        }

        public int[] narr(int n)
        {
            int[] a = new int[n];
            for(int i=0; i<n; ++i)
            {
                a[i] = nint();
            }
            return a;
        }

        public long[] nlarr(int n)
        {
            long[] a = new long[n];
            for(int i=0; i<n; ++i)
            {
                a[i] = nlong();
            }
            return a;
        }
    }


}


