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.*;
import java.math.BigInteger;
public class GFG {
    static class edge{
        int s,d,w;
        public edge(int s,int d,int w)
        {
            this.s=s;
            this.d=d;
            this.w=w;
        }
    }
    
    static class graph{
        int v;
        LinkedList<edge> li[];
        boolean visit[];
        
        public graph(int v)
        {
            this.v=v;
            visit=new boolean[v];
            li=new LinkedList[v];
            for(int i=0;i<v;i++)
            li[i]=new  LinkedList<>();
        }
        public void addedge(int x,int y,int w)
        {
            edge e=new edge(x,y,w);
            li[x].add(e);
            //edge e1=new edge(y,x,w);
            //li[y].add(e1);
        }
            
        public int bfs(int s)
        {
            LinkedList<Integer> q=new LinkedList<Integer>();
            q.add(s);
            int p=0,c=0,p1=0;
            visit[s]=true;
            s=q.poll();
             Iterator<edge> i1=li[s].listIterator();
            while(i1.hasNext()){int n=i1.next().d;p++;visit[n]=true;
            q.add(n);}
            if(p>0)
            c++;
            while(q.size()!=0)
            {
                p1=0;
                while(p>0){
                s=q.poll();
                Iterator<edge> i=li[s].listIterator();
            while(i.hasNext()){int n=i.next().d;
                if(!visit[n]){p1++;visit[n]=true;
                    q.add(n);}
                }p--;}
                p=p1;
                if(p>0)
                c++;
            }
            c++;
            return c;
        }
    }
        public static void main (String[] args) {
		Scanner sc=new Scanner(System.in);
		
        int n=sc.nextInt();
        graph g=new graph(n);
        int min=Integer.MAX_VALUE;
        ArrayList<Integer> arr=new ArrayList<Integer>();
        for(int i=0;i<n;i++)
        {
            int x=sc.nextInt();
            if(x==-1){
            arr.add(i);
            continue;}
            g.addedge(x-1,i,0);
        }
        //System.out.println(min);
        int sum=0;
        for(int i=0;i<arr.size();i++)
        sum=(int)Math.max(sum,g.bfs(arr.get(i)));
        System.out.println(sum);
        }
}