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 arr[100000];
int p(int ind, int c) {
  if (arr[ind] == -1) {
    return c;
  } else {
    return p(arr[ind], c + 1);
  }
}
class Graph {
 private:
  int nodes;
  list<int> *adj;

 public:
  Graph(int nodes) {
    this->nodes = nodes;
    adj = new list<int>[nodes];
  }
  void addEdge(int x, int y) {
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  bool isCyclic(int nodes, bool visited[], int parent) {
    visited[nodes] = true;
    list<int>::iterator i;
    for (i = adj[nodes].begin(); i != adj[nodes].end(); i++) {
      if (!visited[*i]) {
        if (isCyclic(*i, visited, nodes)) {
          return true;
        }
      } else if (*i != parent) {
        return true;
      }
    }
    return false;
  }
  bool isTree() {
    bool *visited = new bool[nodes];
    for (int i = 0; i < nodes; i++) {
      visited[i] = false;
    }
    if (isCyclic(0, visited, -1)) {
      return false;
    }
    for (int u = 0; u < nodes; u++) {
      if (!visited[u]) {
        return false;
      }
    }
    return true;
  }
};
int main() {
  int n;
  cin >> n;
  int maxi = 0;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  for (int i = 1; i <= n; i++) {
    maxi = max(maxi, p(i, 1));
  }
  cout << maxi;
}
