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 tree = std::vector<std::list<int>>;
int height(const tree& organization, int manager = 0) {
  int levels = 0;
  for (const int& employee : organization[manager]) {
    levels = std::max(levels, height(organization, employee));
  }
  return levels + 1;
}
int main(int argc, char const* argv[]) {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int n, parent, k;
  std::cin >> n;
  tree organization(n + 1);
  for (int k = 1; k <= n; k++) {
    std::cin >> parent;
    organization[parent == -1 ? 0 : parent].push_back(k);
  }
  std::cout << height(organization) - 1 << std::endl;
  return 0;
}
