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;
vector<int> e[200000 + 4];
bool vis[200004] = {false};
int level[200005] = {0}, ans = 0;
int n;
void bfs(int a) {
  if (vis[a]) return;
  queue<int> q;
  q.push(a);
  vis[a] = true;
  int sz = 1;
  for (; !q.empty(); sz = q.size()) {
    while (sz--) {
      int v = q.front();
      q.pop();
      vis[v] = true;
      for (int i = 0; i < e[v].size(); i++) {
        if (!vis[e[v][i]]) {
          vis[e[v][i]] = true;
          q.push(e[v][i]);
        }
      }
    }
    ans++;
  }
}
vector<int> ro;
int main() {
  int x;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (x != -1) {
      e[x].push_back(i);
    } else
      ro.push_back(i);
  }
  int ma = -1;
  for (int i = 0; i < ro.size(); i++) {
    ans = 0;
    bfs(ro[i]);
    ma = max(ma, ans);
  }
  cout << ma << endl;
  return 0;
}
