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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class A115 {

    public static class V
    {
        int ind;
        int bigboy;
        List<V> roadsTo = new ArrayList<>();

        public V(int ind){this.ind = ind;}
    }

    public static int deep(V cur, int deep)
    {
        int ret = 0;
        if (cur.roadsTo.size() == 0)return deep;
        for (V kid : cur.roadsTo)
        {
            ret = Math.max(ret, deep(kid, deep + 1));
        }
        return ret;
    }

    public static void main(String... xxx) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        int n = Integer.parseInt(st.nextToken());
        V[] a = new V[n];
        for (int i = 0; i < n; i++) {
            a[i] = new V(i);
        }

        for (int i = 0; i < n; i++) {
            st = new StringTokenizer(br.readLine());
            int boss = Integer.parseInt(st.nextToken());
            if (boss != -1)
            {
                boss--;
                a[boss].roadsTo.add(a[i]);
            }
            else
            {
                a[i].bigboy = 1;
            }
        }
        int ans = 0;
        for (int i = 0; i < n; i++) {
            if (a[i].bigboy == 1)
            {
                ans = Math.max(ans, deep(a[i], 1));
            }
        }

        System.out.println(ans);
        return;
    }
}
