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> adj[2010];
int vis[2010];
int edge_id[2010];
int dist[2010];
int dfs(int s) {
  memset(vis, 0, sizeof(vis));
  memset(edge_id, 0, sizeof(edge_id));
  stack<int> S;
  S.push(s);
  vis[s] = 1;
  dist[s] = 1;
  while (!S.empty()) {
    int u = S.top();
    S.pop();
    while (edge_id[u] < adj[u].size()) {
      int v = adj[u][edge_id[u]];
      edge_id[u]++;
      if (vis[v] == 0) {
        vis[v] = 1;
        dist[v] = dist[u] + 1;
        S.push(u);
        S.push(v);
        break;
      }
    }
  }
  int ret = 0;
  for (int i = 0; i < 2010; i++) {
    if (dist[i] > ret) ret = dist[i];
  }
  return ret;
}
int main() {
  int n, temp, maxs = 0;
  scanf("%d", &n);
  int x[n + 1];
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x[i]);
    if (x[i] != -1) {
      adj[x[i]].push_back(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (x[i] == -1) {
      temp = dfs(i);
      if (temp > maxs) maxs = temp;
    }
  }
  cout << maxs;
}
