Codeforces Beta Round #87 (Div. 2 Only)-Party（DFS找树的深度）

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 nemployees 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

思路：可以把他们的关系看成一棵树，去寻找树的最大深度，就可以用DSF来找

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define rep(i,n) for(int i=1;i<=N;i++)
using namespace std;
 
int a[2005];
int pre[2005];
int sum;
int DFS(int x)
{
	if(pre[x]==x)
	{
		return x;
	}
	else
	{
	  sum++;
//	  cout<<pre[x]<<endl;
	  return DFS(pre[x]);	
	}
}
 
int main()
{
	int N;
	cin>>N;
	int i;
	rep(i,N)
	scanf("%d",&a[i]);
	rep(i,N)
	pre[i]=i;
	rep(i,N)
	{
		if(a[i]!=-1)
		pre[i]=a[i];
		
		else
		{
			pre[i]=i;
		}
		
	}
	int ans=1;
	rep(i,N)
	{
		sum=1;
		DFS(i);
		ans=max(ans,sum);
	}
	cout<<ans<<endl;
	return 0;
}