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<int> vis, lvl, tab;
vector<vector<pair<int, int> > > Adjlist;
int n;
int res = 0;
void bfs(int x) {
  vis[x] = 1;
  lvl[x] = 1;
  res = max(lvl[x], res);
  queue<int> q;
  q.push(x);
  while (!q.empty()) {
    int s = q.front();
    q.pop();
    for (int i = 0; i < (int)Adjlist[s].size(); i++) {
      if (vis[Adjlist[s][i].first] == 0) {
        vis[Adjlist[s][i].first] = 1;
        q.push(Adjlist[s][i].first);
        lvl[Adjlist[s][i].first] = lvl[s] + 1;
        res = max(res, lvl[s] + 1);
      }
    }
  }
}
int main() {
  cin >> n;
  Adjlist.resize(n);
  vis.resize(n);
  lvl.resize(n);
  tab.resize(n);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    tab[i] = x;
    if (x != -1) {
      Adjlist[x - 1].push_back(pair<int, int>(i, 1));
    }
  }
  for (int i = 0; i < n; i++) {
    if (tab[i] == -1) {
      bfs(i);
    }
  }
  cout << res << endl;
  return 0;
}
