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.*;
import java.util.*;
public class Main
{
public static void main(String args[]) throws Exception
{
int graph[][],i,j,n,k,vis[],l,m,root[];
Queue< Integer > Q;
BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
n=Integer.parseInt(in.readLine());
graph=new int[n+3][n+3];
vis=new int[n+3];
root=new int[n+3];
Arrays.fill(root,0);
Arrays.fill(vis,-1);
for(i=0;i<n+2;++i)
Arrays.fill(graph[i],0);
for(i=1;i<=n;++i)
{
j=Integer.parseInt(in.readLine());
if(j!=-1)
graph[j][i]=1;
else
root[i]=1;
}
for(i=1;i<=n;i++)
{
if(vis[i]==-1 && root[i]==1)
{
Q=new LinkedList<Integer>();
Q.add(i);
vis[i]=1;
while(Q.size()!=0)
{
l=Q.poll();
for(m=1;m<=n;m++)
if(graph[l][m]==1 && vis[m]==-1)
{
vis[m]=vis[l]+1;
Q.add(m);
}
}
}
}
k=-1;
for(i=1;i<=n;i++)
{
//System.out.println(vis[i]);
k=Math.max(k,vis[i]);
}
System.out.println(k);
}
}