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;
const long long inf = 1e18;
bool isprime(long long n) {
  for (long long i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
long long factorial(long long n) {
  return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
long long max(long long a, long long b) {
  long long sol = a > b ? a : b;
  return sol;
}
long long min(long long a, long long b) {
  long long sol = a < b ? a : b;
  return sol;
}
vector<long long> v[200005];
long long vis[200005];
long long cnt = 0;
vector<long long> d(1000005, 0);
void dfs(long long node, long long di) {
  vis[node] = 1;
  d[node] = max(d[node], di);
  for (long long i : v[node])
    if (!vis[i]) dfs(i, di + 1);
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long test = 1;
  while (test--) {
    long long n;
    cin >> n;
    long long r = 0;
    for (long long i = 0; i < n; i++) {
      long long x;
      cin >> x;
      if (x > 0) v[x].push_back(i + 1);
    }
    for (long long i = 1; i <= n; i++) {
      memset(vis, 0, sizeof(vis));
      dfs(i, 1);
    }
    cout << *max_element(d.begin(), d.end()) << '\n';
  }
  return 0;
}
