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;
vector<int> tp;
int visited[2111], f[2111], n;
vector<vector<int> > e;
void topo(int u) {
  visited[u] = 1;
  for (int i = 0; i < e[u].size(); i++) {
    int v = e[u][i];
    if (!visited[v]) topo(v);
  }
  tp.push_back(u);
}
void dfs(int u) {
  f[u] = 1;
  int tmp = 0;
  for (int i = 0; i < e[u].size(); i++) {
    int v = e[u][i];
    if (!f[v]) dfs(v);
    tmp = max(tmp, f[v]);
  }
  f[u] += tmp;
}
void process() {
  for (int i = 0; i < n; i++)
    if (visited[i] == 0) topo(i);
  int res = 0;
  reverse(tp.begin(), tp.end());
  for (int i = 0; i < n; i++) {
    int u = tp[i];
    if (!f[u]) {
      dfs(u);
      res = max(res, f[u]);
    }
  }
  cout << res;
}
int main() {
  cin >> n;
  e.assign(n, vector<int>());
  for (int v = 0; v < n; v++) {
    int u;
    scanf("%d", &u);
    if (u != -1) e[u - 1].push_back(v);
  }
  process();
  return 0;
}
