codefoeces 115A Party

点击打开链接

A. Party

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

求树的最大深度，用并查集，不用压缩路径。

#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    int a[2005];
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    int ans=0,temp;
    for(int i=1;i<=n;i++)
    {
        temp=0;
        for(int j=i;j<=n&&j!=-1;j=a[j])
            temp++;
        ans=max(ans,temp);
    }
    printf("%d\n",ans);
    return 0;
}