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;
long long n;
long long p[2003];
long long level[2003];
long long deg[2003];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    cin >> n;
    for (long long i = 0; i < n; ++i) {
      cin >> p[i];
      if (p[i] > 0) {
        p[i]--;
      }
      if (p[i] >= 0) deg[p[i]]++;
    }
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < n; j++) {
        if (deg[j] == 0) {
          deg[j] = 1;
          if (p[j] >= 0) {
            level[p[j]] = max(level[p[j]], level[j] + 1);
            deg[p[j]]--;
          }
        }
      }
    }
    long long ans = 0;
    for (long long i = 0; i < n; ++i) {
      if (level[i] + 1 > ans) {
        ans = level[i] + 1;
      }
    }
    cout << ans;
  }
}
