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;
int par[3000];
int vis[3000];
int bt[3000];
vector<int> adj[3000];
int main() {
  int n, i, k;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> k;
    bt[i] = 0;
    if (k > 0) {
      adj[k].push_back(i);
    }
    if (k == -1) {
      bt[i] = 1;
    }
    vis[i] = 0;
    par[i] = 1;
  }
  int mx = 0;
  for (i = 1; i <= n; i++) {
    if (bt[i] == 1 && vis[i] == 0) {
      vis[i] = 1;
      queue<int> q;
      q.push(i);
      while (!q.empty()) {
        int j = q.front();
        q.pop();
        for (int l = 0; l < adj[j].size(); l++) {
          if (vis[adj[j][l]] == 0) {
            par[adj[j][l]] = 1 + par[j];
            q.push(adj[j][l]);
            vis[adj[j][l]] = 1;
          }
        }
      }
    }
  }
  for (i = 1; i <= n; i++) {
    if (par[i] > mx) {
      mx = par[i];
    }
  }
  cout << mx;
  return 0;
}
