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;
using ll = long long;
ll md = 1000000007;
ll exp(ll a, ll b) {
  ll r = 1ll;
  while (b > 0) {
    if (b & 1) {
      r = r * (a % md);
      r = (r + md) % md;
    }
    b >>= 1;
    a = (a % md) * (a % md);
    a = (a + md) % md;
  }
  return (r + md) % md;
}
ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
T Min(T a, T b) {
  if (a < b) return a;
  return b;
}
template <class T>
T Max(T a, T b) {
  if (a > b) return a;
  return b;
}
const long long Mx = 2e3 + 5;
vector<ll> adj[Mx], depth(Mx, 0), root;
long long n, a;
void dfs(long long x) {
  for (auto& v : adj[x]) dfs(v), depth[x] = Max(depth[x], depth[v]);
  ++depth[x];
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (ll i = 1; i <= n; ++i) {
    cin >> a;
    if (a != -1)
      adj[a].push_back(i);
    else
      root.push_back(i);
  }
  long long ans = 0;
  for (auto& x : root) {
    dfs(x);
    ans = Max(ans, depth[x]);
  }
  cout << ans;
  cout << '\n';
  return 0;
}
