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;
void file() {}
const int MN = 5e4 + 1e2;
const int MW = 1e3 + 5;
const int OO = 1e9 + 5;
int N, x, vs[2005];
vector<int> adj[2005];
int bfs(int s, int cur = 0, int dep = 0, int siz = 0) {
  queue<int> q;
  vs[s] = 1;
  q.push(s);
  for (dep = 1, siz = q.size(); !q.empty(); ++dep, siz = q.size())
    while (siz--) {
      cur = q.front(), q.pop();
      for (auto it : adj[cur])
        if (!vs[it]) vs[it] = 1, q.push(it);
    }
  return dep - 1;
}
int main() {
  file();
  scanf("%d", &N);
  for (int i = 1; i <= N and scanf("%d", &x); ++i)
    adj[(x == -1) ? 0 : x].push_back(i);
  int ret = 0;
  for (auto it : adj[0]) memset(vs, 0, sizeof(vs)), ret = max(ret, bfs(it));
  printf("%d", ret);
  return 0;
}
