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 a {
public static void main(String[] args) throws IOException
{
    //input.init(System.in);
    Scanner input = new Scanner(System.in);
    int n = input.nextInt();
    tot = n;
    data = new int[n];
    for(int i = 0; i<n; i++)
    {
        data[i] = input.nextInt();
    }
    int res = 1;
    for(int i = 0; i<n; i++)
        if(data[i] == -1)
            res = Math.max(res, dfs(i));
    System.out.println(res);
}
public static int[] data;
public static int tot;
public static int dfs(int n)
{
    int res = 0;
    for(int i = 0; i<tot;i++)
    {
        if(data[i] == n+1)
            res = Math.max(res,dfs(i));
    }
    return res+1;
}
public static long gcd(long a, long b)
{
if(b == 0) return a;
return gcd(b, a%b);
}
public static class input {
    static BufferedReader reader;
    static StringTokenizer tokenizer;

    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
        reader = new BufferedReader(
                     new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }

    /** get next word */
    static String next() throws IOException {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                   reader.readLine() );
        }
        return tokenizer.nextToken();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt( next() );
    }
    
    static double nextDouble() throws IOException {
        return Double.parseDouble( next() );
    }
    static long nextLong() throws IOException {
        return Long.parseLong( next() );
    }
}
}
