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;
bool visited[2002];
int d[2002];
vector<int> graph[2002];
int bfs(int start) {
  queue<int> Q;
  Q.push(start);
  memset(visited, false, sizeof(visited));
  memset(d, 0, sizeof(d));
  d[start] = 1;
  visited[start] = true;
  int mx = d[start];
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (unsigned int j = 0; j < graph[u].size(); j++) {
      if (!visited[graph[u][j]]) {
        visited[graph[u][j]] = true;
        d[graph[u][j]] = d[u] + 1;
        mx = max(mx, d[graph[u][j]]);
        Q.push(graph[u][j]);
      }
    }
  }
  return mx;
}
int main() {
  int n, a[2002];
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i + 1];
    if (a[i + 1] != -1) graph[a[i + 1]].push_back(i + 1);
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == -1) {
      int x = bfs(i);
      ans = max(ans, x);
    }
  }
  cout << ans << endl;
  return 0;
}
