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;
template <class T>
T abss(T a) {
  return a > 0 ? a : -a;
}
const auto N = 2000;
deque<int> arr[N + 1];
list<int> tops;
void dfs(int num, int& len) {
  if (arr[num].size() == 0)
    return;
  else {
    ++len;
    list<int> lens_ch;
    for (auto a : arr[num]) {
      int tmp_len = 0;
      dfs(a, tmp_len);
      lens_ch.push_back(tmp_len);
    }
    len += *max_element(lens_ch.begin(), lens_ch.end());
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int num, cnt = 0;
  size_t _q_;
  cin >> _q_;
  while (_q_--) {
    ++cnt;
    cin >> num;
    if (num == -1)
      tops.push_back(cnt);
    else {
      arr[num].push_back(cnt);
    }
  }
  vector<int32_t> lens(tops.size(), 0);
  cnt = 0;
  for (auto top : tops) {
    dfs(top, lens[cnt++]);
  }
  auto max_ = *max_element(lens.begin(), lens.end()) + 1;
  cout << max_ << '\n';
  return 0;
}
