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 double EPS = 1e-8;
const int INF = (int)2e9;
const int MINF = (int)-2e9;
const long long MOD = (long long)1e9 + 7;
const long long LINF = (long long)4e18;
const long long MLINF = (long long)-4e18;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
string to_string(int num) {
  ostringstream ost;
  ost << num;
  return ost.str();
}
bool isSimple(int a) {
  int limit = floor(sqrt((double)a));
  for (int i = 2; i <= limit; i++)
    if (a % i == 0) return false;
  return true;
}
vector<vector<int> > tree;
int n;
vector<bool> used;
int dfs(int v, int count) {
  if (tree[v].size() == 0) return count;
  int Max = MINF;
  for (int i = 0; i < (int)tree[v].size(); i++)
    Max = max(Max, dfs(tree[v][i], count + 1));
  return Max;
}
int main() {
  if (!true) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  }
  cin >> n;
  for (int i = 0; i < n + 1; i++) {
    vector<int> tmp;
    tree.push_back(tmp);
    used.push_back(false);
  }
  for (int i = 1; i <= n; i++) {
    int tmp;
    cin >> tmp;
    if (tmp != -1) tree[tmp].push_back(i);
  }
  int Max = MINF;
  for (int i = 1; i <= n; i++) Max = max(Max, dfs(i, 1));
  cout << Max << endl;
  return 0;
}
