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 Adj_matrix[2009][2009];
int depth[2009];
int status[2009];
void dfs_visit(int v, int n) {
  status[v] = 1;
  int i;
  for (i = 1; i <= n; i++) {
    if (Adj_matrix[v][i] == 1) {
      if (status[i] == 0) dfs_visit(i, n);
      if (depth[i] + 1 > depth[v]) depth[v] = depth[i] + 1;
    }
  }
  status[v] = 2;
}
void dfs(int n) {
  int i;
  for (i = 1; i <= n; i++) {
    status[i] = 0;
    depth[i] = 0;
  }
  for (i = 1; i <= n; i++) {
    if (status[i] == 0) dfs_visit(i, n);
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  cin >> n;
  int A[n], i;
  for (i = 0; i < n; i++) {
    cin >> A[i];
    if (A[i] == -1) continue;
    Adj_matrix[A[i]][i + 1] = 1;
  }
  dfs(n);
  int max = 0;
  for (i = 1; i <= n; i++) {
    if (depth[i] > max) max = depth[i];
  }
  cout << max + 1 << endl;
  return 0;
}
