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;
struct rec {
  int x;
  rec* next;
};
int p[3000];
int v[3000];
rec a[3000];
int main() {
  int i, j, max, n;
  vector<int> q;
  rec* pp;
  scanf("%d", &n);
  for (i = 0; i < n; i++) a[i].next = NULL;
  q.clear();
  memset(v, 0, sizeof(v));
  for (i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    if (p[i] == -1) {
      q.push_back(i);
      v[i] = 1;
    } else {
      p[i]--;
      pp = new rec();
      pp->x = i;
      pp->next = a[p[i]].next;
      a[p[i]].next = pp;
    }
  }
  for (i = 0; i < q.size(); i++) {
    pp = a[q[i]].next;
    while (pp != NULL) {
      if (v[pp->x] == 0) {
        v[pp->x] = v[q[i]] + 1;
        q.push_back(pp->x);
      }
      pp = pp->next;
    }
  }
  max = 0;
  for (i = 0; i < n; i++)
    if (v[i] > max) max = v[i];
  printf("%d\n", max);
  return 0;
}
