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;
long long int n = 0;
vector<vector<long long int> > adj;
vector<bool> visited(10000);
long long int depth[10000];
queue<long long int> q;
void dfs(long long int v) {
  visited[v] = true;
  for (long long int u = 0; u < adj[v].size(); u++) {
    long long int k = adj[v][u];
    if (visited[k] == false) {
      depth[k] = depth[v] + 1;
      dfs(k);
    }
  }
  return;
}
int main() {
  long long int x = 0, j = 0;
  cin >> n;
  adj.resize(n + 1);
  vector<long long int> v(n + 1);
  for (long long int i = 1; i < n + 1; i++) {
    cin >> v[i];
    x = v[i];
    if (x > 0) {
      adj[x].push_back(i);
      adj[i].push_back(x);
    }
  }
  for (long long int i = 1; i < n + 1; i++) depth[i] = 0;
  for (long long int i = 1; i < n + 1; i++) {
    if (v[i] == -1) {
      depth[i] = 1;
      dfs(i);
    }
  }
  for (long long int i = 1; i < n + 1; i++) j = max(j, depth[i]);
  cout << j << endl;
  return 0;
}
