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 INF = 1E9;
long long M = 1E9 + 7;
inline long long max_(long long a, long long b) { return (a > b) ? a : b; }
inline long long min_(long long a, long long b) { return (a < b) ? a : b; }
inline long long power(long long a, long long p) {
  long long res = 1, x = a;
  while (p) {
    if (p & 1) res = (res * x);
    x = (x * x);
    p >>= 1;
  }
  return res;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& p) {
  os << "[ ";
  for (T x : p) os << x << " ";
  os << "]" << endl;
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& p) {
  os << "{ ";
  for (T x : p) os << x << " ";
  os << "}" << endl;
  return os;
}
template <typename Tk, typename Tv>
ostream& operator<<(ostream& os, const map<Tk, Tv>& p) {
  os << "{ ";
  for (pair<Tk, Tv> x : p) os << x << " ";
  os << "}" << endl;
  return os;
}
template <typename Tk, typename Tv>
ostream& operator<<(ostream& os, const pair<Tk, Tv>& p) {
  os << "{" << p.first << ',' << p.second << "}";
  return os;
}
long long N;
vector<long long> adj[2005];
vector<long long> start;
long long ans = 0;
void dfs(long long x, long long len) {
  if (adj[x].size() == 0) ans = max(ans, len);
  long long i;
  for (i = 0; i < adj[x].size(); i++) {
    dfs(adj[x][i], len + 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  if (fopen("inp.txt", "r")) {
    freopen("inp.txt", "r", stdin);
    freopen("outp.txt", "w", stdout);
  }
  cin >> N;
  long long i, p;
  for (i = 1; i <= N; i++) {
    cin >> p;
    if (p == -1) {
      start.push_back(i);
    } else {
      adj[p].push_back(i);
    }
  }
  for (i = 0; i < start.size(); i++) dfs(start[i], 1);
  cout << ans << endl;
  return 0;
}
