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;
}

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