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 mod = 1e9 + 7;
long long expo(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y % 2) res = (res * x % mod) % mod;
    x = (x * x) % mod;
    y /= 2;
  }
  return res;
}
long long inv(long long x) { return expo(x, mod - 2); }
long long add(long long x, long long y) { return (x + y) % mod; }
long long sub(long long x, long long y) { return (x - y) % mod; }
long long mul(long long x, long long y) {
  return ((x % mod) * (y % mod)) % mod;
}
void dfs(vector<vector<long long> > &a, vector<long long> &d, long long st,
         long long n) {
  stack<long long> s;
  s.push(st);
  d[st] = 1;
  vector<bool> v(n + 1, false);
  v[st] = true;
  while (!s.empty()) {
    long long x = s.top();
    s.pop();
    for (long long i = 0; i < a[x].size(); i++) {
      if (!v[a[x][i]]) {
        s.push(a[x][i]);
        v[a[x][i]] = true;
        d[a[x][i]] = d[x] + 1;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> p(n + 1, 0);
  vector<vector<long long> > a(n + 1);
  vector<long long> temp;
  for (long long i = 1; i <= n; i++) {
    cin >> p[i];
    if (p[i] == -1) {
      temp.push_back(i);
      continue;
    }
    a[i].push_back(p[i]);
    a[p[i]].push_back(i);
  }
  vector<long long> d(n + 1, 1);
  for (long long i = 0; i < temp.size(); i++) {
    dfs(a, d, temp[i], n);
  }
  long long ans = -1;
  for (long long i = 1; i < d.size(); i++) ans = max(d[i], ans);
  cout << ans;
  cout << "\n";
}
