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 
#include <bits/stdc++.h>
using namespace std;
int backtracking(vector<vector<int> >& graph, int x) {
  int r = 1;
  for (int i = (int)0; i < (int)graph[x].size(); i++)
    r = max(r, backtracking(graph, graph[x][i]) + 1);
  return r;
}
int main(int argc, char** argv) {
  int emp;
  cin >> emp;
  vector<vector<int> > graph(emp);
  vector<int> supers;
  for (int i = (int)0; i < (int)emp; i++) {
    int a;
    cin >> a;
    if (a == -1)
      supers.push_back(i);
    else
      graph[a - 1].push_back(i);
  }
  int ci = 1;
  for (int i = (int)0; i < (int)supers.size(); i++) {
    int x = supers[i];
    int t = backtracking(graph, x);
    if (t > ci) ci = t;
  }
  cout << ci << endl;
  return 0;
}
