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.*;

public class C116 {
    
    HashSet<Integer> visited;
    int maxDepth;
    
    public void init() {
        visited = new HashSet<Integer>();
        maxDepth = 0;
    }
    
    public int getMaxDepth() {
        return maxDepth;
    }
    
    public void DFS(boolean a[][], int i, int depth) {
        maxDepth = Math.max(depth, maxDepth);
        visited.add(i);
        for (int j = 0; j < a[i].length; j++) {
            if (!visited.contains(j) && a[i][j])
                DFS(a, j, depth + 1);
        }
    }
    
    public static void main(String args[]) {
        C116 obj = new C116();
        obj.init();
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        boolean graph[][] = new boolean[n + 1][n + 1];
        for (int i = 1; i <= n; i++) {
            int p = in.nextInt();
            if (p == -1)
                graph[0][i] = true;
            else
                graph[p][i] = true;
        }
        obj.DFS(graph, 0, 0);
        System.out.println(obj.getMaxDepth());
    }
}
