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;
int row[] = {0, 1, 0, -1};
int col[] = {1, 0, -1, 0};
const long long N = 2e3 + 10;
const long long M = 1e18 + 7;
const long long H = 1e9 + 9;
const long long base = 131;
long long logx(long long a, long long bs) { return log(a) / log(bs); }
long long Max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long Min(long long a, long long b) {
  if (a > b) return b;
  return a;
}
long long big_mul(long long a, long long b, long long m) {
  long long x = 0, y = a % m;
  while (b > 0) {
    if (b & 1) x = (x + y) % m;
    y = (2 * y) % m;
    b >>= 1;
  }
  return x % m;
}
long long big_mod(long long a, long long b, long long m) {
  long long x = 1, y = a % m;
  while (b > 0) {
    if (b & 1) x = (x * y) % m;
    y = (y * y) % m;
    b >>= 1;
  }
  return x % m;
}
long long Big_mod(long long a, long long b, long long m) {
  long long x = 1, y = a % m;
  while (b > 0) {
    if (b & 1) x = big_mul(x, y, m);
    y = big_mul(y, y, m);
    b >>= 1;
  }
  return x % m;
}
long long mod_inverse(long long a, long long m) { return big_mod(a, m - 2, m); }
long long GCD(long long a, long long b) {
  if (!a) return b;
  return GCD(b % a, a);
}
vector<bool> vis(N);
vector<int> boss;
vector<int> level(N);
vector<vector<int> > vec(N);
void dfs(int x, int cnt) {
  vis[x] = true;
  level[x] = cnt;
  for (int i = 0; i < vec[x].size(); i++) {
    int p = vec[x][i];
    if (vis[p] == false) dfs(p, cnt + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    if (a > 0)
      vec[a].push_back(i + 1);
    else
      boss.push_back(i + 1);
  }
  for (int i = 0; i < boss.size(); i++) {
    int p = boss[i];
    if (vis[p] == false) dfs(p, 1);
  }
  int ans = 0;
  for (int i = 1; i < n + 1; i++) ans = max(ans, level[i]);
  cout << ans << endl;
  return 0;
}
