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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
unsigned seed = std::chrono::system_clock::now().time_since_epoch().count();
using namespace std;
template <class T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &a : v) is >> a;
  return is;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (const T &t : v) os << t << " ";
  return os << endl;
}
template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &v) {
  return os << v.first << " " << v.second;
}
double gcd(double a, double b) { return a < 0.01 ? b : gcd(fmod(b, a), a); }
long long bpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long int kv;
long long int t, n, x, y, h, k, z, p, q, w, m;
string s, s1, s2;
long long int dp[100000];
vector<long long int> v;
long long int sg(long long int s) {
  if (dp[s] != 0) return dp[s];
  if (v[s - 1] == -1) return dp[s] = 1;
  return dp[s] = sg(v[s - 1]) + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(16);
  ;
  cin >> n;
  v.resize(n);
  cin >> v;
  kv = 1;
  for (long long i = 0; i < n; i++) {
    kv = max(kv, sg(i + 1));
  }
  cout << kv;
  return 0;
}
