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<vector<long long> > v;
vector<long long> p;
long long bfs(long long src) {
  queue<long long> q;
  q.push(src);
  long long h = 0;
  while (!q.empty()) {
    long long x = q.size();
    while (x--) {
      long long z = q.front();
      q.pop();
      for (long long i = 0; i < v[z].size(); i++) {
        q.push(v[z][i]);
      }
    }
    h++;
  }
  return h;
}
int main() {
  std::ios::sync_with_stdio(false);
  long long t = 1;
  while (t--) {
    long long n, x;
    cin >> n;
    v.resize(n + 1);
    for (long long i = 1; i <= n; i++) {
      cin >> x;
      if (x != -1) {
        v[x].push_back(i);
      } else
        p.push_back(i);
    }
    long long ans = 0;
    for (long long i = 0; i < p.size(); i++) {
      ans = max(ans, bfs(p[i]));
    }
    cout << ans << "\n";
  }
  return 0;
}
