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 Party {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] direction = new int[n+1];
        int[] maxDist = new int[n+1];
        boolean[] isSup = new boolean[n+1];
        int curNode, temp;
        LinkedList<Integer> parents = new LinkedList<Integer>(), bfs = new LinkedList<Integer>();
        for(int i = 1; i <= n; ++i) {
            direction[i] = in.nextInt();
            if(direction[i] != -1)
                isSup[direction[i]] = true;
        }
        for(int i = 1; i <= n; ++i) {
            if(isSup[i])
                continue;
            else {
                parents.add(i);
            }
        }
        
        while(parents.peek() != null) {
            curNode = parents.poll();
            if(direction[curNode] == -1)
                continue;
            else {
                temp = direction[curNode];
                if(maxDist[temp] < (maxDist[curNode] + 1)) {
                    maxDist[temp] = maxDist[curNode]+1;
                    bfs.add(temp);
                    while(bfs.peek() != null) {
                        curNode = bfs.poll();
                        if(direction[curNode] == -1)
                            continue;
                        else {
                            temp = direction[curNode];
                            if(maxDist[temp] < (maxDist[curNode] + 1)) {
                                maxDist[temp] = maxDist[curNode] + 1;
                                bfs.add(temp);
                            }
                        }
                    }
                }
            }
        }
        int max = 0;
        for(int i = 1; i <= n; ++i) {
            if(maxDist[i] > max)
                max = maxDist[i];
        }
        System.out.println(max+1);
    }
}