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<vector<int> > adj;
vector<bool> visited;
vector<int> level;
int z = 0;
void bfs(int s) {
  queue<int> queue;
  queue.push(s);
  if (level[s] == 0) level[s] = 1;
  while (!queue.empty()) {
    s = queue.front();
    queue.pop();
    for (int i = 0; i < adj[s].size(); i++) {
      int child = adj[s][i];
      level[child] = level[s] + 1;
      queue.push(child);
    }
  }
}
void dfs(int v, int c) {
  z = max(z, c);
  if (visited[v]) return;
  visited[v] = true;
  for (int i = 0; i < adj[v].size(); i++) {
    int child = adj[v][i];
    if (!visited[child]) {
      dfs(child, c + 1);
    }
  }
}
int main() {
  int n, z;
  cin >> n;
  adj = vector<vector<int> >(n + 1);
  level = vector<int>(n + 1);
  visited = vector<bool>(n + 1, false);
  vector<int> qrr(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> z;
    if (z != -1) {
      adj[z].push_back(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    bfs(i);
  }
  int m = *max_element(level.begin(), level.end());
  cout << m << endl;
  return 0;
}
