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.awt.*;
import java.util.LinkedList;
import java.util.Scanner;

public class Main {

    static class Forest {
        private int n;
        private int mDepth;
        private LinkedList<Point>[] vertices;
        private boolean[] visited;
        Forest (int n) {
            this.vertices = new LinkedList[n];
            this.visited = new boolean[n];
            this.mDepth = 0;
            for (int i = 0; i < n; i++) {
                vertices[i] = new LinkedList<Point>();
                visited[i] = false;
            }
        }
        // directed.
        void connect (Point p ,Point q) {
            this.vertices[p.x].addFirst(q);
        }
        void computeDepth (Point root) {
            // base case.
            if (vertices[root.x].isEmpty()) {
                mDepth = Math.max(mDepth ,root.y);
                return;
            }
            visited[root.x] = true;
            for (int i = 0; i < vertices[root.x].size(); i++) {
                Point v = vertices[root.x].get(i);
                if (!visited[v.x]) {
                    v.y = root.y + 1;
                    vertices[root.x].set(i ,v);
                    computeDepth(v);
                }
            }
        }
    }
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        int noOfGroups = 0;
        Forest f = new Forest(n);
        LinkedList<Point> roots = new LinkedList<Point>();
        for (int i = 0; i < n; i++) {
            int p = input.nextInt();
            if (p == -1)
                roots.addFirst(new Point(i ,0));
            else    f.connect(new Point((p-1) ,0) ,new Point(i ,0));
        }
        for (int i = 0; i < roots.size(); i++) {
            f.computeDepth(roots.get(i));
            noOfGroups = Math.max(noOfGroups ,f.mDepth);
            f.mDepth = 0;
        }
        System.out.println(noOfGroups + 1);
    }
}
