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;
vector<vector<long> > tr;
vector<long> v;
long bfs(long r) {
  v[r] = 1;
  long i, hight = 1;
  queue<pair<long, long> > q;
  pair<long, long> u;
  q.push(make_pair(r, 1));
  while (!q.empty()) {
    u = q.front();
    q.pop();
    long no = u.first, ht = u.second;
    for (i = 0; i < tr[no].size(); i++) {
      if (v[tr[no][i]] == 0) {
        v[tr[no][i]] = 1;
        if (ht + 1 > hight) hight = ht + 1;
        q.push(make_pair(tr[no][i], ht + 1));
      }
    }
  }
  return hight;
}
int main() {
  long i, j, k, n;
  cin >> n;
  vector<long> roots;
  tr.resize(n + 1);
  for (i = 1; i <= n; i++) {
    cin >> j;
    if (j == -1)
      roots.push_back(i);
    else
      tr[j].push_back(i);
  }
  long ans = -50000;
  v.resize(n + 1);
  for (i = 1; i <= n; i++) v[i] = 0;
  for (i = 0; i < roots.size(); i++) {
    ans = max(ans, bfs(roots[i]));
  }
  cout << ans << endl;
  return 0;
}
