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 N = 2005;
const long double PI = 3.1415926535;
const long double eps = 1e-07;
const long long INF = 4.1e18;
const long long inf = 1e9;
const long long M = 305;
const string IO[2]{"NO\n", "YES\n"};
const string Io[2]{"No\n", "Yes\n"};
const string io[2]{"no\n", "yes\n"};
vector<long long> a[N], d(N, 0);
void dfs(long long v, long long dd = 1) {
  d[v] = dd;
  for (auto u : a[v]) {
    dfs(u, dd + 1);
  }
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> s(N, 1);
  for (long long i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    if (x == -1) continue;
    --x;
    s[i] = 0;
    a[x].push_back(i);
  }
  for (long long i = 0; i < n; ++i) {
    if (s[i]) {
      dfs(i);
    }
  }
  long long ans = 1;
  for (long long i = 0; i < n; ++i) {
    ans = max(ans, d[i]);
  }
  cout << ans << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t = 1;
  while (t--) solve();
}
