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 int fast_exp(long long int a, long long int b) {
  long long int res = 1;
  res %= 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
    }
    b = b >> 1;
    a = (a * a) % 1000000007;
  }
  return res;
}
int gcd(int a, int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
vector<int> factorize(int n) {
  vector<int> res;
  for (int i = 2; i * i <= n; ++i) {
    while (n % i == 0) {
      res.push_back(i);
      n /= i;
    }
  }
  if (n != 1) {
    res.push_back(n);
  }
  return res;
}
int col[4000];
vector<vector<int> > adj(3000);
int dfs(int v) {
  col[v] = 1;
  int x = 1;
  for (int i = 0; i < adj[v].size(); i++) {
    { x = max(x, 1 + dfs(adj[v][i])); }
  }
  return x;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  int i, x;
  memset(col, -1, sizeof(col));
  for (i = 1; i < n + 1; i++) {
    cin >> x;
    if (x != -1) {
      adj[x].push_back(i);
    }
  }
  int cnt = 0;
  for (i = 1; i < n + 1; i++) {
    if (col[i] == -1) {
      cnt = max(cnt, dfs(i));
    }
  }
  cout << cnt << endl;
}
