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;
map<int, bool> visited;
int bfs(int s, int n, vector<int> adj[]) {
  queue<int> visit;
  visit.push(s);
  visited[s] = true;
  int ht = 1, h = 1;
  map<int, int> lvl;
  lvl[s] = 1;
  while (!visit.empty()) {
    int v = visit.front();
    visit.pop();
    if (!adj[v].size()) continue;
    for (int u : adj[v]) {
      if (!visited[u]) {
        visited[u] = true;
        visit.push(u);
        lvl[u] = lvl[v] + 1;
        h = max(h, lvl[u]);
      }
    }
  }
  return h;
}
int main() {
  int n, x;
  cin >> n;
  vector<int> adj[n + 1], tr;
  visited.clear();
  for (int i = 1; i != n + 1; i += 1) {
    cin >> x;
    if (x == -1) {
      tr.push_back(i);
      continue;
    }
    adj[x].push_back(i);
    adj[i].push_back(x);
  }
  int height = 1;
  for (int au : tr) {
    visited.clear();
    height = max(height, bfs(au, n, adj));
  }
  cout << height << endl;
  return 0;
}
