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;
const int64_t inf = 1e18;
vector<vector<int64_t> > adj;
int64_t inp() {
  int64_t x;
  cin >> x;
  return x;
}
vector<int64_t> level;
int64_t res = 0;
void dfs(int64_t u, int64_t LV = 0) {
  level[u] = LV;
  res = max(res, LV);
  for (auto v : adj[u]) {
    dfs(v, LV + 1);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  for (int64_t _ = (1); _ >= (1); _--) {
    int64_t n;
    cin >> n;
    adj.resize(n + 1);
    level.resize(n + 1);
    for (int64_t node = (1); node <= (n); node++)
      adj[max(inp(), int64_t(0))].push_back(node);
    dfs(0);
    cout << res;
  }
  return 0;
}
