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 pi = 3.14159265358979323846;
long long mod = 1000000007;
int mul(int x, int y) { return (x * 1ll * y) % mod; }
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res % mod;
}
int inv(int x) { return power(x, mod - 2); }
int divide(int x, int y) { return mul(x, inv(y)); }
int fact[200001];
void precalc() {
  fact[0] = 1;
  for (int i = 1; i < 200001; i++) fact[i] = mul(fact[i - 1], i);
}
int C(int n, int k) { return divide(fact[n], mul(fact[k], fact[n - k])); }
const int dx[] = {0, 1, 0, -1, 1, -1, 1, -1};
const int dy[] = {1, 0, -1, 0, 1, -1, -1, 1};
int vis[2001];
;
vector<int> level(2001);
vector<int> adj[2001];
void bfs(int s) {
  queue<int> q;
  q.push(s);
  level[s] = 1;
  vis[s] = true;
  while (!q.empty()) {
    int p = q.front();
    q.pop();
    for (int i = 0; i < adj[p].size(); i++) {
      if (vis[adj[p][i]] == false) {
        level[adj[p][i]] = level[p] + 1;
        q.push(adj[p][i]);
        vis[adj[p][i]] = true;
      }
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed;
  cout.precision(12);
  int q = 1;
  while (q--) {
    int n;
    cin >> n;
    int p[n + 1];
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      if (x != -1) adj[x].push_back(i);
      p[i] = x;
    }
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < 2001; j++) vis[j] = level[j] = 0;
      if (p[i] == -1) bfs(i);
      ans = max(ans, *max_element(level.begin(), level.end()));
    }
    cout << ans << endl;
  }
  return 0;
}
