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;
const double pi = 3.14159265358979323846;
void cp() {}
long long getfather(long long a[], long long n) {
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long j = i, count = 0;
    while (a[j] != -1) {
      count++;
      j = a[j];
    }
    ans = max(ans, count);
  }
  return ans + 1;
}
void solve() {
  long long n;
  cin >> n;
  long long a[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  cout << getfather(a, n) << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long t = 1;
  while (t--) solve();
  return 0;
}
