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

Output

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;

}

巴特西

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

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