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;
int mod = 1000000007;
const int N = 2005;
vector<int> v[N];
bool visited[N];
int bfs(int s) {
  int level[N];
  memset(level, 0, sizeof(level));
  level[s] = 1;
  queue<int> q;
  q.push(s);
  visited[s] = 1;
  int ans = 1;
  while (!q.empty()) {
    s = q.front();
    q.pop();
    ans = max(ans, level[s]);
    for (int i = 0; i < v[s].size(); i++) {
      if (!visited[v[s][i]]) {
        visited[v[s][i]] = 1;
        q.push(v[s][i]);
        level[v[s][i]] = level[s] + 1;
      }
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> st;
  int ans = 1;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (x == -1)
      st.push_back(i + 1);
    else
      v[x].push_back(i + 1);
  }
  for (int i = 0; i < st.size(); i++) {
    ans = max(ans, bfs(st[i]));
  }
  cout << ans;
  return 0;
}
