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;
map<int, vector<int> > graph;
int visited[2001];
int cost[2001];
int dfs(int start, int end) {
  int ret = 0;
  visited[start] = 0;
  for (int i = 0; i < graph[start].size(); i++) {
    int no = graph[start][i];
    if (visited[no] == -1) {
      cost[no] += cost[start] + 1;
      dfs(no, end);
    }
  }
  if (start == end) return 0;
  return cost[end];
}
int main() {
  int n;
  cin >> n;
  vector<int> inis;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    if (x > 0) {
      graph[x].push_back(i);
      graph[i].push_back(x);
    } else {
      inis.push_back(i);
    }
  }
  memset(visited, -1, sizeof visited);
  memset(cost, 0, sizeof cost);
  int m = -1;
  for (vector<int>::iterator itr = inis.begin(); itr != inis.end(); ++itr) {
    if (*itr == n)
      dfs(*itr, 1);
    else
      dfs(*itr, n);
    for (int i = 1; i <= n; ++i) {
      m = max(m, cost[i]);
    }
  }
  cout << m + 1 << endl;
  return 0;
}
