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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  vector<long long int> v[20000], v1;
  long long int deg[20000] = {0};
  map<long long int, long long int> m;
  for (long long int i = 1; i < n + 1; i++) {
    long long int a;
    cin >> a;
    if (a != -1) {
      v[a].push_back(i);
    } else {
      v1.push_back(i);
    }
  }
  long long int visited[20000];
  memset(visited, 0, sizeof(visited));
  long long int ans = 1;
  queue<long long int> q;
  for (long long int i = 0; i < v1.size(); i++) {
    q.push(v1[i]);
    deg[v1[i]] = 1;
    while (!q.empty()) {
      long long int p = q.front();
      visited[p] = 1;
      q.pop();
      for (long long int j = 0; j < v[p].size(); j++) {
        if (visited[v[p][j]] == 0) {
          deg[v[p][j]] = deg[p] + 1;
          q.push(v[p][j]);
        }
      }
    }
  }
  for (long long int i = 0; i < 20000; i++) {
    ans = max(ans, deg[i]);
  }
  cout << ans << endl;
  return 0;
}
