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.PrintStream;
import java.util.ArrayList;
import java.util.Scanner;

public class Main {
    private PrintStream out;
    private Scanner in;

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        new Main(in, System.out).solve();
        in.close();
    }

    public Main(Scanner in, PrintStream out) {
        this.in = in;
        this.out = out;
    }

    public void solve() {
        // TODO: Solve the problem
        int n = in.nextInt();
        ArrayList<Integer>[] childs = new ArrayList[n];
        ArrayList<Integer> roots = new ArrayList<Integer>();

        for(int i=0; i<n; ++i) {
            childs[i] = new ArrayList<Integer>();
        }

        for(int i=0; i<n; ++i) {
            int x = in.nextInt();
            if( x == -1 ) {
                roots.add(i);
            } else {
                x--;
                childs[x].add(i);
            }
        }

        int mx = 0;
        for(Integer root : roots) {
            int nDepth = depth(root, childs);
            if( nDepth > mx ) {
                mx = nDepth;
            }
        }

        out.println(mx);
    }

    public int depth(int root, ArrayList<Integer>[] childs) {
        ArrayList<Integer> rootChilds = childs[root];

        int mx = 1;

        for(Integer child : rootChilds) {
            int ndepth = 1 + depth(child, childs);
            if(ndepth > mx) {
                mx = ndepth;
            }
        }

        return mx;
    }
}
