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 
//Codeforces 115A 
import java.util.Scanner;
import java.util.Vector;



public class CF115A {
    static final Scanner SC = new Scanner(System.in);
    public static void main(String[] args) {
        int employees = SC.nextInt();
        int[] managedBy = new int[employees];
        for (int e = 0; e < employees; ++e)
            managedBy[e] = SC.nextInt();
        DFS hierarchy = new DFS(employees, managedBy);
        int groups = hierarchy.getMaxDepth();
        System.out.println(groups);
    }
}

class DFS {
    int vertices;
    Vector<Integer>[] connected;
    boolean[] hasParent; // Notes whether or not an employee is managed by another
    public DFS(int vertices, int[] edges) {
        this.vertices = vertices;
        hasParent = new boolean[vertices];
        this.connected = (Vector<Integer>[]) new Vector[vertices];
        for (int v = 0; v < vertices; ++v)
            this.connected[v] = new Vector<Integer>();
        for (int v = 0; v < vertices; ++v)
            if(edges[v] != -1) {
                connected[v].add(edges[v]-1);
                connected[edges[v]-1].add(v);
                hasParent[v] = true;
            }
    }

    // Computes and returns maximum depth in hierarchy
    public int getMaxDepth() {
        boolean[] visited = new boolean[vertices];
        int maxDepth = 0;
        for (int v = 0; v < vertices; ++v)
            if (!hasParent[v]) // If employee v is not managed we check the depth of branch
                maxDepth = Math.max(maxDepth, depthFrom(v, visited));
        return maxDepth;
    }

    // Computes and returns depth in hierarchy with 'v' as superior-most employee
    public int depthFrom(int v, boolean[] visited) {
        visited[v] = true;
        int depth = 1;
        for (int i = 0; i < connected[v].size(); ++i)
            if (!visited[connected[v].get(i)])
                depth = Math.max(depth, 1 + depthFrom(connected[v].get(i), visited));
        return depth;
    }
}