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 int MAXN = 1e5 + 7;
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const long long MINF = 1e18 + 7;
const int EPS = 1e-6;
const int dx[] = {0, 0, 1, -1, -1, -1, 1, 1};
const int dy[] = {1, -1, 0, 0, -1, 1, -1, 1};
const double pi = acos(-1.0);
long long a[MAXN], b[MAXN];
vector<int> g[MAXN], v;
bool used[MAXN];
long long n, m, k, cnt = 0, sum = 0, ans = 0, mx = 0, mn = INF;
bool y = 0;
void dfs(int v, int stol) {
  b[v] = stol;
  used[v] = 1;
  mx = max(b[v], mx);
  for (int i = 0; i < g[v].size(); i++) {
    if (!used[g[v][i]]) {
      dfs(g[v][i], stol + 1);
    }
  }
}
void freopen() {
  freopen("A.in", "r", stdin);
  freopen("A.out", "w", stdout);
}
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] != -1) {
      g[a[i]].push_back(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] == -1) {
      dfs(i, 1);
    }
  }
  cout << mx;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  long long T = 1;
  while (T--) {
    solve();
  }
  return 0;
}
